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  • I can't insert data into my database

    - by Ken
    I don't know why, but my data doesn't go into my database 'users' with the table 'data'. <html> <body> <?php date_default_timezone_set("America/Los_Angeles"); include("mainmenu.php"); $con = mysql_connect("localhost", "root", "g00dfor@boy"); if(!$con){ die(mysql_error()); } $usrname = $_POST['usrname']; $fname = $_POST['fname']; $lname = $_POST['lname']; $password = $_POST['password']; $email = $_POST['email']; mysql_select_db(`users`, $con) or die(mysql_error()); mysql_query(INSERT INTO `users`.`data` (`id`, `usrname`, `fname`, `lname`, `email`, `password`) VALUES (NULL, '$usrname', '$fname', '$lname', '$email', 'password')) or die(mysql_error()); mysql_close($con) echo("Thank you for registering!"); ?> </body> </html> All i get is a blank page.

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  • sending +-200 emails using php mail() function in a loop

    - by Glenn
    Note: It is worth noting that the mail() function is not suitable for larger volumes of email in a loop. This function opens and closes an SMTP socket for each email, which is not very efficient. Source: PHP manual What are larger volumes? A 100 or a 1000?? Can I safely make it loop 200 times without much problems? (I can't install pear)

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  • htmlspecialchars() - How and when to use and avoid multiple use

    - by naescent
    Hi, I'm building a PHP intranet for my boss. A simple customer, order, quote system. It will be denied access from the Internet and only used by 3 people. I'm not so concerned with security as I am with validation. Javascript is disables on all machines. The problem I have is this: Employee enters valid data into a form containing any of the following :;[]"' etc. Form $_POSTS this data to a validationAndProcessing.php page, and determines whether the employee entered data or not in to the fields. If they didn't they are redirected back to the data input page and the field they missed out is highlighted in red. htmlspecialchars() is applied to all data being re-populated to the form from what they entered earlier. Form is then resubmitted to validationAndProcessing.php page, if successful data is entered into the database and employee is taken to display data page. My question is this: If an employee repeatedly enters no data in step 1, they will keep moving between step 1 and 4 each time having htmlspecialchars() applied to the data. So that:- & becomes:- &amp; becomes:- &amp;amp; becomes:- &amp;amp;amp; etc.. How can I stop htmlspecialchars() being applied multiple times to data that is already cleaned? Thanks, Adam

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  • PHP PDO fetch null

    - by Jacob
    How do you check if a columns value is null? Example code: $db = DBCxn::getCxn(); $sql = "SELECT exercise_id, author_id, submission, result, submission_time, total_rating_votes, total_rating_values FROM submissions LEFT OUTER JOIN submission_ratings ON submissions.exercise_id=submission_ratings.exercise_id WHERE id=:id"; $st = $db->prepare($sql); $st->bindParam(":id", $this->id, PDO::PARAM_INT); $st->execute(); $row = $st->fetch(); if($this->total_rating_votes == null) // this doesn't seem to work even though there is no record in submission_ratings???? { ... }

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  • Overriding unique indexed values

    - by Yeti
    This is what I'm doing right now (name is UNIQUE): SELECT * FROM fruits WHERE name='apple'; Check if the query returned any result. If yes, don't do anything. If no, a new value has to be inserted: INSERT INTO fruits (name) VALUES ('apple'); Instead of the above is it ok to insert the value into the table without checking if it already exists? If the name already exists in the table, an error will be thrown and if it doesn't, a new record will be inserted. Right now I am having to insert 500 records in a for loop, which results in 1000 queries. Will it be ok to skip the "already-exists" check?

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  • 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?

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  • PHP - Too many mysql_query("SELECT .. ") .. ?

    - by Mike
    Hey, I'm making an e-shop and to display the tree of categories and all the products with their multiple variations of prices I made like more than 150 mysql_query("SELECT ..."); queries on one page. (If I count the "while" loops). Is it too many, and if yes, can it have any negative effect? (ofc. it takes longer to load the data ..) Also can I anyhow achieve the effect of this code without doing it that way? $result2 = mysql_query("SELECT * FROM ceny WHERE produkt_id='$id' ORDER BY gramaz"); $result3 = mysql_query("SELECT * FROM ceny WHERE produkt_id='$id' ORDER BY gramaz"); $result4 = mysql_query("SELECT * FROM ceny WHERE produkt_id='$id' ORDER BY gramaz"); $result5 = mysql_query("SELECT * FROM ceny WHERE produkt_id='$id' ORDER BY gramaz"); while( $row2 = mysql_fetch_array( $result2 )) { } while( $row3 = mysql_fetch_array( $result2 )) { } while( $row4 = mysql_fetch_array( $result2 )) { } while( $row5 = mysql_fetch_array( $result2 )) { } Thanks, Mike.

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  • Dynamic evaluation of a table column within an insert before trigger

    - by Tim Garver
    HI All, I have 3 tables, main, types and linked. main has an id column and 32 type columns. types has id, type linked has id, main_id, type_id I want to create an insert before trigger on the main table. It needs to compare its 32 type columns to the values in the types table if the main table column has an 'X' for its value and insert the main_id and types_id into the linked table. i have done a lot of searching, and it looks like a prepared statement would be the way to go, but i wanted to ask the experts. The issue, is i dont want to write 32 IF statements, and even if i did, i need to query the types table to get the ID for that type, seems like a huge waist of resources. Ideally i want to do this inside of my trigger: BEGIN DECLARE @types results_set -- (not sure if this is a valid type); -- (iam sure my loop syntax is all wrong here)... SET @types = (select * from types) for i=0;i<types.records;i++ { IF NEW.[i.type] = 'X' THEN insert into linked (main_id,type_id) values (new.ID, i.id); END IF; } END; Anyway, This is what i was hoping to do, maybe there is a way to dynamically set the field name inside of a results loop, but i cant find a good example of this. Thanks in advance Tim

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  • How to build a SQL statement when any combination of user input to the table is possible?

    - by Greg McNulty
    Example: the user fills in everything but the product name. I need to search on what is supplied, so in this case everything but productName= This example could be for any combination of input. Is there a way to do this? Thanks. $name = $_POST['n']; $cat = $_POST['c']; $price = $_POST['p']; if( !($name) ) { $name = some character to select all? } $sql = "SELECT * FROM products WHERE productCategory='$cat' and productName='$name' and productPrice='$price' "; EDIT Solution does not have to protect from attacks. Specifically looking at the dynamic part of it.

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  • How to do left joins with least-n-per-group query?

    - by Nate
    I'm trying to get a somewhat complicated query working and am not having any luck whatsoever. Suppose I have the following tables: cart_items: +--------------------------------------------+ | item_id | cart_id | movie_name | quantity | +--------------------------------------------+ | 0 | 0 | braveheart | 4 | | 1 | 0 | braveheart | 9 | | . | . | . | . | | . | . | . | . | | . | . | . | . | | . | . | . | . | +--------------------------------------------+ movies: +------------------------------+ | movie_id | movie_name | ... | +------------------------------+ | 0 | braveheart | . | | . | . | . | | . | . | . | | . | . | . | | . | . | . | +------------------------------+ pricing: +-----------------------------------------+ | id | movie_name | quantity | price_per | +-----------------------------------------+ | 0 | braveheart | 1 | 1.99 | | 1 | braveheart | 2 | 1.50 | | 2 | braveheart | 4 | 1.25 | | 3 | braveheart | 8 | 1.00 | | . | . | . | . | | . | . | . | . | | . | . | . | . | | . | . | . | . | | . | . | . | . | +-----------------------------------------+ I need to join the data from the tables, but with the added complexity that I need to get appropriate price_per from the pricing table. Only one price should be returned for each cart_item, and that should be the lowest price from the pricing table where the quantity for the cart item is at least the quantity in the pricing table. So, the query should return for each item in cart_items the following: +---------------------------------------------+ | item_id | movie_name | quantity | price_per | +---------------------------------------------+ Example 1: Variable passed to the query: cart_id = 0. Return: +---------------------------------------------+ | item_id | movie_name | quantity | price_per | +---------------------------------------------+ | 0 | braveheart | 4 | 1.25 | | 1 | braveheart | 9 | 1.00 | +---------------------------------------------+ Note that this is a minimalist example and that additional data will be pulled from the tables mentioned (particularly the movies table). How could this query be composed? I have tried using left joins and subqueries, but the difficult part is getting the price and nothing I have tried has worked. Thanks for your help. EDIT: I think this is similar to what I have working with my "real" tables: SELECT t1.item_id, t2.movie_name, t1.quantity FROM cart_items t1 LEFT JOIN movies t2 ON t2.movie_name = t1.movie_name WHERE t1.cart_id = 0 Assuming I wrote that correctly (I quickly tried to "port over" my real query), then the output would currently be: +---------------------------------+ | item_id | movie_name | quantity | +---------------------------------+ | 0 | braveheart | 4 | | 1 | braveheart | 9 | +---------------------------------+ The trouble I'm having is joining the price at a certain quantity for a movie. I simply cannot figure out how to do it.

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  • INSERT INTO table doesn't work???

    - by Joann
    I found a tutorial from nettuts and it has a source code in it so tried implementing it in my site.. It is working now. However, it doesn't have a Registration system so I am making one. The thing is, as I have expected, my code is not working... It doesn't seem to know how to INSERT into the database. Here's the function that inserts data into the db. function register_User($un, $email, $pwd) { $query = "INSERT INTO users( username, password, email ) VALUES(:uname, :pwd, :email) LIMIT 1"; if($stmt = $this->conn->prepare($query)) { $stmt->bind_param(':uname', $un); $stmt->bind_param(':pwd', $pwd); $stmt->bind_param(':email', $email); $stmt->execute(); if($stmt->fetch()) { $stmt->close(); return true; } else return "The username or email you entered is already in use..."; } } I have debugged the connection to the database from within the class, it says it's connected. I tried using this method instead: function register($un, $email, $pwd) { $registerquery = $this->conn->query( "INSERT INTO users(uername, password, email) VALUES('".$un."', '".$pwd."', '".$email."')"); if($registerquery) { echo "<h4>Success</h4>"; } else { echo "<h4>Error</h4>"; } } And it echos "Error"... Can you please help me pen point the error in this??? :(

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  • 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?

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  • How to retrieve column total when rows are paginated?

    - by Rick
    Hey guys I have a column "price" in a table and I used a pagination script to generate the data displayed. Now the pagination is working perfectly however I am trying to have a final row in my HTML table to show the total of all the price. So I wrote a script to do just that with a foreach loop and it sort of works where it does give me the total of all the price summed up together however it is the sum of all the rows, even the ones that are on following pages. How can I retrieve just the sum of the rows displayed within the pagination? Thank you! Here is the query.. SELECT purchase_log.id, purchase_log.date_purchased, purchase_log.total_cost, purchase_log.payment_status, cart_contents.product_name, members.first_name, members.last_name, members.email FROM purchase_log LEFT JOIN cart_contents ON purchase_log.id = cart_contents.purchase_id LEFT JOIN members ON purchase_log.member_id = members.id GROUP BY id ORDER BY id DESC LIMIT 0,30";

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  • Combine SQL statement

    - by ninumedia
    I have 3 tables (follows, postings, users) follows has 2 fields - profile_id , following_id postings has 3 fields - post_id, profile_id, content users has 3 fields - profile_id, first_name, last_name I have a follows.profile_id value of 1 that I want to match against. When I run the SQL statement below I get the 1st step in obtaining the correct data. However, I now want to match the postings.profile_id of this resulting set against the users table so each of the names (first and last name) are displayed as well for all the listed postings. Thank you for your help! :) Ex: SELECT * FROM follows JOIN postings ON follows.following_id = postings.profile_id WHERE follows.profile_id = 1

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  • Zend_Table_Db and Zend_Paginator num rows

    - by Uffo
    I have the following query: $this->select() ->where("`name` LIKE ?",'%'.mysql_escape_string($name).'%') Now I have the Zend_Paginator code: $paginator = new Zend_Paginator( // $d is an instance of Zend_Db_Select new Zend_Paginator_Adapter_DbSelect($d) ); $paginator->getAdapter()->setRowCount(200); $paginator->setItemCountPerPage(15) ->setPageRange(10) ->setCurrentPageNumber($pag); $this->view->data = $paginator; As you see I'm passing the data to the view using $this->view->data = $paginator Before I didn't had $paginator->getAdapter()->setRowCount(200);I could determinate If I have any data or not, what I mean with data, if the query has some results, so If the query has some results I show the to the user, if not, I need to show them a message(No results!) But in this moment I don't know how can I determinate this, since count($paginator) doesn't work anymore because of $paginator->getAdapter()->setRowCount(200);and I'm using this because it taks about 7 sec for Zend_Paginator to count the page numbers. So how can I find If my query has any results?

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  • How to insert <br/> after each 5 results?

    - by Axel
    This is my code: $query = mysql_query("SELECT * FROM books ORDER BY id") or die(mysql_error()); while($row = mysql_fetch_assoc($query)) { echo $row["bookname"]." - "; } How to make only 5 books displayed in each line, by inserting a at the start if the row is 5 or 10 or 15 etc... Thanks

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  • Magento - Data is not inserted into database, but the id is autoincremented

    - by Joseph
    I am working on a new payment module for Magento and have come across an issue that I cannot explain. The following code that runs after the credit card is verified: $table_prefix = Mage::getConfig()->getTablePrefix(); $tableName = $table_prefix.'authorizecim_magento_id_link'; $resource = Mage::getSingleton('core/resource'); $writeconnection = $resource->getConnection('core_write'); $acPI = $this->_an_customerProfileId; $acAI = $this->_an_customerAddressId; $acPPI = $this->_an_customerPaymentProfileId; $sql = "insert into {$tableName} values ('', '$customerId', '$acPI', '$acPI', '3')"; $writeconnection->query($sql); $sql = "insert into {$tableName} (magCID, anCID, anOID, anObjectType) values ('$customerId', '$acPI', '$acAI', '2')"; $writeconnection->query($sql); $sql = "insert into {$tableName} (magCID, anCID, anOID, anObjectType) values ('$customerId', '$acPI', '$acPPI', '1')"; $writeconnection->query($sql); I have verified using Firebug and FirePHP that the SQL queries are syntactically correct and no errors are returned. The odd thing here is that I have checked the database, and the autoincrement value is incremented on every run of the code. However, no rows are inserted in the database. I have verified this by adding a die(); statement directly after the first write. Any ideas why this would be occuring? The relative portion of the config.xml is this: <config> <global> <models> <authorizecim> <class>CPAP_AuthorizeCim_Model</class> </authorizecim> <authorizecim_mysql4> <class>CPAP_AuthorizeCim_Model_Mysql4</class> <entities> <anlink> <table>authorizecim_magento_id_link</table> </anlink> </entities> <entities> <antypes> <table>authorizecim_magento_types</table> </antypes> </entities> </authorizecim_mysql4> </models> <resources> <authorizecim_setup> <setup> <module>CPAP_AuthorizeCim</module> <class>CPAP_AuthorizeCim_Model_Resource_Mysql4_Setup</class> </setup> <connection> <use>core_setup</use> </connection> </authorizecim_setup> <authorizecim_write> <connection> <use>core_write</use> </connection> </authorizecim_write> <authorizecim_read> <connection> <use>core_read</use> </connection> </authorizecim_read> </resources> </global> </config>

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  • 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!

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  • Designing a recipe database that needs to include ingredients as well as sub-recipes

    - by VinceL
    I am designing a recipe database that needs to be very flexible as it is going to be communicating directly with our back-of-house inventory system. This is what I have so far in regards to the tables: Recipe: this table will contain the recipe date: the name, steps needed to cook, etc. Ingredients/Inventory: this is our back of house inventory, so this will have the information about each product that will be used in our recipes. Recipe Line Item: This is the tricky table, I want to be able link to the ingredients here as well as the quantity needed for the recipe, but I also need to be able to directly include recipes from the recipe table (such as marinara sauce that we make in-house), and that is why I am having trouble figuring out the best way to design this table. Basically, the recipe line item table needs to be able to link to either the ingredients table or the recipe table depending on which line item is needed and I want to know what would be the most effective way to handle that. Thank you so much in advance!

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