Search Results

Search found 20931 results on 838 pages for 'mysql insert'.

Page 399/838 | < Previous Page | 395 396 397 398 399 400 401 402 403 404 405 406  | Next Page >

  • White Label Ecommerce app. Shared or Individual dbs

    - by MetaDan
    Currently I'm working with an in house white label cms that we resell to multiple clients and it all runs from the same box/db. I'm just looking at converting this to have an ecommerce version that we'll run alongside it. I'm wondering whether there will be an issue keeping all the products/categories/orders in one db or whether it would be advisory to separate each instance of the site into its own db for this. These white label instances will only be sold to smaller companies that probably wont have masses of traffic/products and are looking for a simple ecommerce site. Anything larger will definitely get its own hosting and db. But for smaller scale stuff do you think a single db will be ok?

    Read the article

  • How to compare a string with an option value

    - by user225269
    I have this html form which has options: <tr> <td width="30" height="35"><font size="3">*List:</td> <td width="30"><input name="specific" type="text" id="specific" maxlength="25" value=""> </td> <td><font size="3">*By:</td> <td> <select name="general" id="general"> <font size="3"> <option value="YEAR">Year</option> <option value="ADDRESS">Address</option> </select></td></td> </tr> And I'm trying to have this as the form action: if ('{$_POST["ADDRESS"]}'="ADDRESS") Which will compare if the value in the option in the html form matches the word "Address". If it matches then it will execute this query: $saddress= mysql_real_escape_string($_POST['specific']);<--this is the input form where the user will put the specific address to search. mysql_query("SELECT * FROM student WHERE ADDRESS='$saddress'"); Please I need help in here, I thinks its wrong: if ('{$_POST["ADDRESS"]}'="ADDRESS")

    Read the article

  • Find a date between start_date and end_date

    - by Margaret
    I have a table of events with a recorded start and end time. I want to find all events that occur on a specific date. Some events started a year ago and some will continue farther ahead. I would like to be able to pick for example May 20, 2010 and find all events occurring on that date.

    Read the article

  • PHP Show all data in column

    - by user342391
    I am trying to show all of the data in the 'status' column of my table but am having troubles. What am I doing wrong: <?php $query1 = "SELECT id, status FROM alerts WHERE customerid='".$_SESSION['customerid']."' ORDER BY id LIMIT $start, $limit "; $result = mysql_query($query1); while ($row = mysql_fetch_array($result)) { echo $row['status'] ; } ?>

    Read the article

  • mod rewrite, title slugs and htaccess

    - by chris
    I have been taken in to provide some seo guidance on a website which has been running since 2005. My problem is i want to use clean urls. The code that handles the url is hidden away in some class file.. and with over a few thousand lines of code its a struggle to rewrite it. So I'm think, I have gone through all the products and created a slug for them as a field in the product table. Is it possible to do something like an intermediate file for htaccess. Some thing like 1./clean-slug-comes-in/ 2.htaccess catches this and uses slug.php to find the relevant product id for the slug. 3.Then product.php?id=(ID.found.from.2) is loaded?

    Read the article

  • Product Name Print Several times, How to fix.?

    - by mans
    i had added the following Opencart module for my order report list... http://www.opencart.com/index.php?route=extension/extension/info&extension_id=3597&filter_search=order%20list%20filter%20model&page=4 I have problems with the column "Products". If there are more than one option the products name prints several times. So if I got a product with three options the product name prints three times. Is there any way to fix this problem? i want print product name and model number only once, any idea.? i will attach the results what i got now... this is my sql query... public function getOrders($data = array()) { $sql = "select o.order_id,o.email,o.telephone,CONCAT(o.shipping_address_1, ' ', o.shipping_address_2) AS address,CONCAT(o.firstname, ' ', o.lastname) AS customer,o.payment_zone AS state,o.payment_address_2 AS block, o.payment_address_1 AS address,o.payment_postcode AS postcode,(SELECT os.name FROM " . DB_PREFIX . "order_status os WHERE os.order_status_id = o.order_status_id AND os.language_id = '" . (int)$this->config->get('config_language_id') . "') AS status,o.payment_city AS city,GROUP_CONCAT(pd.name) AS pdtname,GROUP_CONCAT(op.model) AS model,o.date_added,sum(op.quantity) AS quantity,GROUP_CONCAT(opt.value ) AS options, GROUP_CONCAT(opt.order_product_id ) AS ordprdid,GROUP_CONCAT(op.order_product_id ) AS optprdid, GROUP_CONCAT(op.quantity) AS opquantity from `" . DB_PREFIX . "order` o LEFT JOIN " . DB_PREFIX . "order_product op ON (op.order_id = o.order_id) LEFT JOIN " . DB_PREFIX . "product_description pd ON (pd.product_id = op.product_id and pd.language_id = '" . (int)$this->config->get('config_language_id') . "') LEFT JOIN " . DB_PREFIX . "order_option opt ON (opt.order_product_id = op.order_product_id) "; Product Name = GROUP_CONCAT(pd.name) AS pdtname,

    Read the article

  • Making a relevant search of text in database using regex

    - by madphp
    Can anyone tell me how I could count the possible instances of a keyword in a block of text? I've split a search term up into separate tokens, so just need to run through and do a count for every instance and removing punctuation or other special characters when making the count. Secondly, if someone has inserted search terms surrounded by double quotes, i want to be able to skip explode, but just count instances of that exact phrase. It doesn't have to be case sensitive and I would like to remove punctuation from the phrase when doing the count. Thirdly, in both cases i want to be able to ignore wordpress and html tags. Lastly, if anyone know any good tutorials for relevant searches that answer the questions above, that would cool too. I've got this far. $results = $wpdb->get_results($sql); $tokens = explode('search_terms'); // Re-arrange Relevant Results foreach ($results As $forum_topic){ foreach($tokens As $token){ // count tokens in topic_title if ($token ){ } } }

    Read the article

  • How to insert custom date time in oracle using java?

    - by shree
    Hi i have a column (type date).I want to insert custom date and time without using Preparedstatement .i have used String date = sf.format(Calendar.getInstance().getTime()); String query = "Insert into entryTbl(name, joinedDate, ..etc) values ("abc", to_date(date, 'yyyy/mm/dd HH:mm:ss'))"; statement.executeUpdate(query); but am getting literal doesnot match error. so even tried with "SYSDATE".Its inserting only date not time.So how to insert the datetime using java into oracle?please any one help..

    Read the article

  • How to construct this query? (Ordering by COUNT() and joining with users table)

    - by Andrew
    users table: id-user-other columns scores table: id-user_id-score-other columns They're are more than one rows for each user, but there's only two scores you can have. (0 or 1, == win or loss). So I want to output all the users ordered by the number of wins, and all the users ordered by the numbers of losses. I know how to do this by looping through each user, but I was wondering how to do it with one query. Any help is appreciated!

    Read the article

  • Object of class mysqli_result could not be converted to string

    - by Joann
    I asked Google to help me I got no luck. :-( Here's the particular code that generates the error: $this->conn->query("UPDATE tz_members SET confirm='yes' WHERE usr='".$uname."'"); The whole function is the following: function update_confirm_field($code) { $uname = $this->conn->query("SELECT usr FROM tz_members WHERE confirm='".$code."'"); $this->conn->query("UPDATE tz_members SET confirm='yes' WHERE usr='".$uname."'"); } Forgive me if I have missed something stupid. Can anyone tell me what's causing the problem please???

    Read the article

  • PHP Multiple User Login Form - Navigation to Different Pages Based on Login Credentials

    - by Zulu Irminger
    I am trying to create a login page that will send the user to a different index.php page based on their login credentials. For example, should a user with the "IT Technician" role log in, they will be sent to "index.php", and if a user with the "Student" role log in, they will be sent to the "student/index.php" page. I can't see what's wrong with my code, but it's not working... I'm getting the "wrong login credentials" message every time I press the login button. My code for the user login page is here: <?php session_start(); if (isset($_SESSION["manager"])) { header("location: http://www.zuluirminger.com/SchoolAdmin/index.php"); exit(); } ?> <?php if (isset($_POST["username"]) && isset($_POST["password"]) && isset($_POST["role"])) { $manager = preg_replace('#[^A-Za-z0-9]#i', '', $_POST["username"]); $password = preg_replace('#[^A-Za-z0-9]#i', '', $_POST["password"]); $role = preg_replace('#[^A-Za-z0-9]#i', '', $_POST["role"]); include "adminscripts/connect_to_mysql.php"; $sql = mysql_query("SELECT id FROM Users WHERE username='$manager' AND password='$password' AND role='$role' LIMIT 1"); $existCount = mysql_num_rows($sql); if (($existCount == 1) && ($role == 'IT Technician')) { while ($row = mysql_fetch_array($sql)) { $id = $row["id"]; } $_SESSION["id"] = $id; $_SESSION["manager"] = $manager; $_SESSION["password"] = $password; $_SESSION["role"] = $role; header("location: http://www.zuluirminger.com/SchoolAdmin/index.php"); } else { echo 'Your login details were incorrect. Please try again <a href="http://www.zuluirminger.com/SchoolAdmin/index.php">here</a>'; exit(); } } ?> <?php if (isset($_POST["username"]) && isset($_POST["password"]) && isset($_POST["role"])) { $manager = preg_replace('#[^A-Za-z0-9]#i', '', $_POST["username"]); $password = preg_replace('#[^A-Za-z0-9]#i', '', $_POST["password"]); $role = preg_replace('#[^A-Za-z0-9]#i', '', $_POST["role"]); include "adminscripts/connect_to_mysql.php"; $sql = mysql_query("SELECT id FROM Users WHERE username='$manager' AND password='$password' AND role='$role' LIMIT 1"); $existCount = mysql_num_rows($sql); if (($existCount == 1) && ($role == 'Student')) { while ($row = mysql_fetch_array($sql)) { $id = $row["id"]; } $_SESSION["id"] = $id; $_SESSION["manager"] = $manager; $_SESSION["password"] = $password; $_SESSION["role"] = $role; header("location: http://www.zuluirminger.com/SchoolAdmin/student/index.php"); } else { echo 'Your login details were incorrect. Please try again <a href="http://www.zuluirminger.com/SchoolAdmin/index.php">here</a>'; exit(); } } ?> And the form that the data is pulled from is shown here: <form id="LoginForm" name="LoginForm" method="post" action="http://www.zuluirminger.com/SchoolAdmin/user_login.php"> User Name:<br /> <input type="text" name="username" id="username" size="50" /><br /> <br /> Password:<br /> <input type="password" name="password" id="password" size="50" /><br /> <br /> Log in as: <select name="role" id="role"> <option value="">...</option> <option value="Head">Head</option> <option value="Deputy Head">Deputy Head</option> <option value="IT Technician">IT Technician</option> <option value="Pastoral Care">Pastoral Care</option> <option value="Bursar">Bursar</option> <option value="Secretary">Secretary</option> <option value="Housemaster">Housemaster</option> <option value="Teacher">Teacher</option> <option value="Tutor">Tutor</option> <option value="Sanatorium Staff">Sanatorium Staff</option> <option value="Kitchen Staff">Kitchen Staff</option> <option value="Parent">Parent</option> <option value="Student">Student</option> </select><br /> <br /> <input type="submit" name = "button" id="button" value="Log In" onclick="javascript:return validateLoginForm();" /> </h3> </form> Once logged in (and should the correct page be loaded, the validation code I have at the top of the script looks like this: <?php session_start(); if (!isset($_SESSION["manager"])) { header("location: http://www.zuluirminger.com/SchoolAdmin/user_login.php"); exit(); } $managerID = preg_replace('#[^0-9]#i', '', $_SESSION["id"]); $manager = preg_replace('#[^A-Za-z0-9]#i', '', $_SESSION["manager"]); $password = preg_replace('#[^A-Za-z0-9]#i', '', $_SESSION["password"]); $role = preg_replace('#[^A-Za-z0-9]#i', '', $_SESSION["role"]); include "adminscripts/connect_to_mysql.php"; $sql = mysql_query("SELECT id FROM Users WHERE username='$manager' AND password='$password' AND role='$role' LIMIT 1"); $existCount = mysql_num_rows($sql); if ($existCount == 0) { header("location: http://www.zuluirminger.com/SchoolAdmin/index.php"); exit(); } ?> Just so you're aware, the database table has the following fields: id, username, password and role. Any help would be greatly appreciated! Many thanks, Zulu

    Read the article

  • Syncing a table records with a Service response frequently

    - by Karthik Dheeraj
    I am requesting data from a service whose response in stored in a database.First, I have an empty table, whenever I make my very first request the records from the service comes to my database table. from now, whenever I make second request, the service will provide me some records which may be same as my first response, may be new records, may be updated records etc. my query is to how to update my table with respect to the responses coming from the service during my second request on-wards? so that Unchanged records will remain same, New records will be added, updated records will be updated.Do I need to write any stored procedure on my DB or any workaround ?what might be the scenario if I use Nomysql DB's like mongo DB ? Thanks In Advance.

    Read the article

  • Can I do this with just SQL?

    - by Josh
    At the moment I have two tables, products and options. Products contains id title description Options contains id product_id sku title Sample data may be: Products id: 1 title: 'test' description: 'my description' Options id: 1 product_id: 1 sku: 1001 title: 'red' id: 2 product_id: 1 sku: 1002 title: 'blue' I need to display each item, with each different option. At the moment, I select the rows in products and iterate through them, and for each one select the appropriate rows from options. I then create an array, similar to: [product_title] = 'test'; [description] = 'my description'; [options][] = 1, 1001, 'red'; [options][] = 2, 1002, 'blue'; Is there a better way to do this with just sql (I'm using codeigniter, and would ideally like to use the Active Record class)?

    Read the article

  • Sample/Example needed for a table/field setup

    - by acctman
    Can someone explain the statement below to me with a working sample/example. thanks in advance. You can not create duplicate fields, but simply add a single extra field, "coupleId", which would have a unique id for each couple; and two rows (one for each person) per couple; then JOIN the table against itself with a constraint like a.coupleId = b.coupleId AND a.id < b.id so that you can condense the data into a single result row for a given couple.

    Read the article

  • Letting users try your web app before sign-up: sessions or temp db?

    - by Mat
    I've seen a few instances now where web applications are letting try them out without you having to sign-up (though to save you need to of course). example: try at http://minutedock.com/ I'm wondering about doing this for my own web app and the fundamental question is whether to store their info into sessions or into a temp user table? The temp user table would allow logging and potentially be less of a hit on the server, correct? Is there a best practice here?

    Read the article

  • change password code error.....

    - by shimaTun
    I've created a code to change a password. Now it seem contain an error.before i fill the form. the page display the error message: Parse error: parse error, unexpected $end in C:\Program Files\xampp\htdocs\e-Complaint(FYP)\userChangePass.php on line 222 this the code: <?php # userChangePass.php //this page allows logged in user to change their password. $page_title='Change Your Password'; //if no first_name variable exists, redirect the user if(!isset($_SESSION['nameuser'])){ header("Location: http://" .$_SERVER['HTTP_HOST']. dirname($_SERVER['PHP_SELF'])."/index.php"); ob_end_clean(); exit(); }else{ if(isset($_POST['submit'])) {//handle form. require_once('connectioncomplaint.php'); //connec to the database //check for a new password and match againts the confirmed password. if(eregi ("^[[:alnum:]]{4,20}$", stripslashes(trim($_POST['password1'])))){ if($_POST['password1'] == $_POST['password2']){ $p =escape_data($_POST['password1']); }else{ $p=FALSE; echo'<p><font color="red" size="+1"> Your password did not match the confirmed password!</font></p>'; } }else{ $p=FALSE; echo'<p><font color="red" size="+1"> Please Enter a valid password!</font></p>'; } if($p){ //if everything OK. //make the query $query="UPDATE access SET password=PASSWORD('$p') WHERE userid={$_SESSION['userid']}"; $result=@mysql_query($query);//run the query. if(mysql_affected_rows() == 1) {//if it run ok. //send an email,if desired. echo '<p><b>your password has been changed.</b></p>'; //include('templates/footer.inc');//include the HTML footer. exit(); }else{//if it did not run ok $message= '<p>Your password could not be change due to a system error.We apolpgize for any inconvenience.</p><p>' .mysql_error() .'</p>'; } mysql_close();//close the database connection. }else{//failed the validation test. echo '<p><font color="red" size="+1"> Please try again.</font></p>'; } }//end of the main Submit conditional. ?> And code for form: <h1>Change Your Password</h1> <form action="<?php echo $_SERVER['PHP_SELF']; ?>" method="post"> <fieldset> <p><b>New Password:</b><input type="password" name="password1" size="20" maxlength="20" /> <small>Use only letters and numbers.Must be between 4 and 20 characters long.</small></p> <p><b>Confirm New Password:</b><input type="password" name="password2" size="20" maxlength="20" /></p> </fieldset> <div align="center"> <input type="submit" name="submit" value="Change My Password" /></div> </form><!--End Form-->

    Read the article

  • change password code error

    - by ejah85
    I've created a code to change a password. Now it seem contain an error. When I fill in the form to change password, and click save the error message: Warning: mysql_real_escape_string() expects parameter 2 to be resource, null given in C:\Program Files\xampp\htdocs\e-Complaint(FYP)\userChangePass.php on line 103 Warning: mysql_real_escape_string() expects parameter 2 to be resource, null given in C:\Program Files\xampp\htdocs\e-Complaint(FYP)\userChangePass.php on line 103 I really don’t know what the error message means. Please guys. Help me fix it. Here's is the code: <?php session_start(); ?> <?php # change password.php //set the page title and include the html header. $page_title = 'Change Your Password'; //include('templates/header.inc'); if(isset($_POST['submit'])){//handle the form require_once('connectioncomplaint.php');//connect to the db. //include "connectioncomplaint.php"; //create a function for escaping the data. function escape_data($data){ global $dbc;//need the connection. if(ini_get('magic_quotes_gpc')){ $data=stripslashes($data); } return mysql_real_escape_string($data, $dbc); }//end function $message=NULL;//create the empty new variable. //check for a username if(empty($_POST['userid'])){ $u=FALSE; $message .='<p> You forgot enter your userid!</p>'; }else{ $u=escape_data($_POST['userid']); } //check for existing password if(empty($_POST['password'])){ $p=FALSE; $message .='<p>You forgot to enter your existing password!</p>'; }else{ $p=escape_data($_POST['password']); } //check for a password and match againts the comfirmed password. if(empty($_POST['password1'])) { $np=FALSE; $message .='<p> you forgot to enter your new password!</p>'; }else{ if($_POST['password1'] == $_POST['password2']){ $np=escape_data($_POST['password1']); }else{ $np=FALSE; $message .='<p> your new password did not match the confirmed new password!</p>'; } } if($u && $p && $np){//if everything's ok. $query="SELECT userid FROM access WHERE (userid='$u' AND password=PASSWORD('$p'))"; $result=@mysql_query($query); $num=mysql_num_rows($result); if($num == 1){ $row=mysql_fetch_array($result, MYSQL_NUM); //make the query $query="UPDATE access SET password=PASSWORD('$np') WHERE userid=$row[0]"; $result=@mysql_query($query);//run the query. if(mysql_affected_rows() == 1) {//if it run ok. //send an email,if desired. echo '<p><b>your password has been changed.</b></p>'; include('templates/footer.inc');//include the HTML footer. exit();//quit the script. }else{//if it did not run OK. $message= '<p>Your password could not be change due to a system error.We apolpgize for any inconvenience.</p><p>' .mysql_error() .'</p>'; } }else{ $message= '<p> Your username and password do not match our records.</p>'; } mysql_close();//close the database connection. }else{ $message .='<p>Please try again.</p>'; } }//end oh=f the submit conditional. //print the error message if there is one. if(isset($message)){ echo'<font color="red">' , $message, '</font>'; } ?> <form action="<?php echo $_SERVER['PHP_SELF']; ?>" method="post"> <body> <script language="JavaScript1.2">mmLoadMenus();</script> <table width="604" height="599" border="0" align="center" cellpadding="0" cellspacing="0"> <tr> <td height="130" colspan="7"><img src="images/banner(E-Complaint)-.jpg" width="759" height="130" /></td> </tr> <tr> <td width="100" height="30" bgcolor="#ABD519"></td> <td width="100" bgcolor="#ABD519"></td> <td width="100" bgcolor="#ABD519"></td> <td width="100" bgcolor="#ABD519"></td> <td width="100" bgcolor="#ABD519"></td> <td width="160" bgcolor="#ABD519"> <?php include "header.php"; ?>&nbsp;</td> </tr> <tr> <td colspan="7" bgcolor="#FFFFFF"> <fieldset><legend> Enter your information in the form below:</legend> <p><b>User ID:</b> <input type="text" name="username" size="10" maxlength="20" value="<?php if(isset($_POST['userid'])) echo $_POST['userid']; ?>" /></p> <p><b>Current Password:</b> <input type="password" name="password" size="20" maxlength="20" /></p> <p><b>New Password:</b> <input type="password" name="password1" size="20" maxlength="20" /></p> <p><b>Confirm New Password:</b> <input type="password" name="password2" size="20" maxlength="20" /></p> </fieldset> <div align="center"> <input type="submit" name="submit" value="Change My Password" /></div> </form><!--End Form--> </td> </tr> </table> </body> </html>

    Read the article

  • Problem updating values in combobox in vb.net

    - by user225269
    I have this code, but I have a problem. When I update but do not really made any changes to the value and press the update button, the data becomes null. And it will seem that I deleted the value. I've taught of a solution, that is to add both combobox1.selectedtext and combobox1.selecteditem to the function. But it doesn't work. combobox1.selecteditem is working when you try to alter the values when you update. But will save a null value when you don't alter the values using the combobox combobox1.selectedtext will save the data into the database even without altering. But will not save the data if you try to alter it. -And I incorporated both of them, but still only one is performing, and I think it is the one that I added first: Dim shikai As New Updater Try shikai.id = TextBox1.Text shikai.fname = TextBox2.Text shikai.mi = TextBox3.Text shikai.lname = TextBox4.Text shikai.ad = TextBox5.Text shikai.contact = TextBox9.Text shikai.year = ComboBox1.SelectedText shikai.section = ComboBox2.SelectedText shikai.gender = ComboBox3.SelectedText shikai.religion = ComboBox4.SelectedText shikai.year = ComboBox1.SelectedItem shikai.section = ComboBox2.SelectedItem shikai.gender = ComboBox3.SelectedItem shikai.religion = ComboBox4.SelectedItem shikai.bday = TextBox6.Text shikai.updates() MsgBox("Successfully updated!") Please help, what would be a simple workaround to solve this problem?

    Read the article

  • Slope requires a real as parameter 2?

    - by Dave Jarvis
    Question How do you pass the correct value to udf_slope's second parameter type? Attempts CAST(Y.YEAR AS FLOAT), but that failed (SQL error). Y.YEAR + 0.0, but that failed, too (see error message). slope(D.AMOUNT, 1.0), failed as well Error Message Using udf_slope fails due to: Can't initialize function 'slope'; slope() requires a real as parameter 2 Code SELECT D.AMOUNT, Y.YEAR, slope(D.AMOUNT, Y.YEAR + 0.0) as SLOPE, intercept(D.AMOUNT, Y.YEAR + 0.0) as INTERCEPT FROM YEAR_REF Y, DAILY D Here, D.AMOUNT is a FLOAT and Y.YEAR is an INTEGER. Create Function The slope function was created as follows: CREATE AGGREGATE FUNCTION slope RETURNS REAL SONAME 'udf_slope.so'; Function Signature From udf_slope.cc: double slope( UDF_INIT* initid, UDF_ARGS* args, char* is_null, char* is_error ) Example Usages Reading the fine manual reveals: UDF intercept() Calculates the intercept of the linear regression of two sets of variables. Function name intercept Input parameter(s) 2 (dependent variable: REAL, independent variable: REAL) Examples SELECT intercept(income,age) FROM customers UDF slope() Calculates the slope of the linear regression of two sets of variables. Function name slope Input parameter(s) 2 (dependent variable: REAL, independent variable: REAL) Examples SELECT slope(income,age) FROM customers Thoughts? Thank you!

    Read the article

< Previous Page | 395 396 397 398 399 400 401 402 403 404 405 406  | Next Page >