Search Results

Search found 29702 results on 1189 pages for 'select insert'.

Page 549/1189 | < Previous Page | 545 546 547 548 549 550 551 552 553 554 555 556  | Next Page >

  • Disabling Text field with Javascript when value in drop down box is from mysql

    - by SteveJ313
    Hi I have a simple script in HTML, using a dropdown menu. When the value 1 is selected, the user can write in the text field, if value 2 is selected, it disables the text field. However, i changed the values of the dropdown menu, so that one value was from a mysql table(using PHP) and the other remained 'option value='1''. Yet now neither text field is disabled. Below is the code. `<script type="text/javascript"> function findselected() { if (document.form.selmenu.value == <?php echo $id; ?>) { document.form.txtField.disabled=true; // return false; // not sure this line is needed } else { document.form.txtField.disabled=false; // return false; // not sure this line is needed } } ` And the PHP section if(mysql_num_rows($SQL) == 1) { echo "<select name='selmenu' onChange='findselected()'>"; echo "<label>TCA_Subject</label>"; while ($row=mysql_fetch_array($SQL)) { echo "<option value='$id'>$thing</option>"; echo "<option value='2'>Choice 2</option>"; } } echo "<option value=$userid>'Choice 1'</option>"; ?> <option value='2'>Choice 2</option>"; </select> I have tried taking the second option value out of the loop, putting it into html, editing the variable in the javascript function. There is not a fault with the PHP as it is retrieving the right results and displaying it, yet the text field doesnt become disabled. Does anyone know of a possible solution? Thanks

    Read the article

  • Not Inserted into database in PHP script

    - by Aruna
    Hi, i am having an form for uploading an excel file like <form enctype="multipart/form-data" action="http://localhost/joomla/Joomla_1.5.7/index.php?option=com_jumi&fileid=7" method="POST"> Please choose a file: <input name="file" type="file" id="file" /><br /> <input type="submit" value="Upload" /> </form> And in the FIle http://localhost/joomla/Joomla_1.5.7/index.php?option=com_jumi&fileid=7 i have retrived the uploaded file contents by <?php echo "Name". $_FILES['file']['name'] . "<br />"; echo "Type: " . $_FILES["file"]["type"] . "<br />"; echo "Size: " . ($_FILES["file"]["size"] / 1024) . " Kb<br />"; echo "Stored in: " . $_FILES["file"]["tmp_name"] . "<br />"; $string = file_get_contents( $_FILES["file"]["tmp_name"] ); foreach ( explode( "\n", $string ) as $userString ) { $userString = trim( $userString ); echo $userString; $array = explode( ';', $userString ); $userdata = array ( 'cf_id'=>trim( $array[0] ), 'text_0'=>trim( $array[1] ), 'text_1'=>trim( $array[2] ), 'text_2'=>trim( $array[3] ), 'text_3'=>trim($array[4]), 'text_6'=>trim($array[5]), 'text_7'=>trim($array[6]), 'text_9'=>trim($array[7]), 'text_12'=>trim($array[8])); global $db; $db =& JFactory::getDBO(); $query = 'INSERT INTO #__chronoforms_UploadAuthor VALUES ("'.$userdata['cf_id'].'","'.$userdata['text_0'].'","'.$userdata['text_1'].'","'.$userdata['text_2'].'","'.$userdata['text_3'].'","'.$userdata['text_6'].'","'.$userdata['text_7'].'","'.$userdata['text_9'].'")'; $db->Execute($query); echo "<br>"; } i am trying to insert the contents into the table #__chronoforms_UploadAuthor in Joomla database . it doent shows any error but it is not inserted into the database.. Please help me.. Y the contents are not inserted into the database...And how to make it inserted into the database..

    Read the article

  • MYSQL autoincrement a column or just have an integer, difference?

    - by David19801
    Hi, If I have a column, set as primary index, and set as INT. If I don't set it as auto increment and just insert random integers which are unique into it, does that slow down future queries compared to autincrementing? Does it speed things up if I run OPTIMIZE on a table with its primary and only index as INT? (assuming only 2 columns, and second column is just some INT value) (the main worry is the upper limit on the autoincrement as theres lots of adds and deletes in my table)

    Read the article

  • Add new Formelemts and transform it with jqtransform

    - by user545782
    I'm just trying to change my formfields with jqtransform. With the following Javascript Code, i add new formfields: <script type="text/javascript"> $(function() { var scntDiv = $('#p_scents'); var i = $('#p_scents p').size() + 1; $('#addScnt').live('click', function() { if(i < 4){ $('<p>\n\ <label for="p_scnts">\n\ <input type="text" id="p_scnt" size="17" name="fmac' + i +'[]" value="" />\n\ <select name="fmac' + i +'[]" size="1" id="p_scnt_a">\n\ <option selected value="K">KABEL</option>\n\ <option value="W">WLAN</option>\n\ <option value="V">VPN</option>\n\ </select>\n\ </label>\n\ </p>').appendTo(scntDiv); i++; } if(i > 3 ){ $('#showaddmac').hide(); } return false; }); $('#remScnt').live('click', function() { if( i > 2 ) { $(this).parents('p').remove(); i--; } if(i < 4 ){ $('#showaddmac').show(); } return false; }); }); </script> this works with out problemes, but the new fields will be not transformed. Does anyone know a solution to the problem? Sry for my bad english :)

    Read the article

  • header location won't work in php

    - by Jayden Kelly
    I am making a login page for my website but the header location won't work. here is the code of login.php: <?php include ( './includes/header.php' ); if (isset($_POST['submit'])) { $username = $_POST['username']; $password = $_POST['password']; $check_username = mysql_query("SELECT username FROM users WHERE username='$username'"); $numrows = mysql_num_rows($check_username); if ($numrows != 1) { echo 'That User doesn\'t exist.'; } else { $check_password = mysql_query("SELECT password FROM users WHERE password='$password' && username='$username'"); while ($row = mysql_fetch_assoc($check_password)) { $password_db = $row['password']; if ($password_db == $password) { $_SESSION['username'] = $username; header("Location: members.php"); } } } } ?> <h2>Login to Your Account</h2> <form action='login.php' method='POST'> <input type='text' name='username' value='Username ...' onclick='value=""'/><p /> <input type='password' name='password' value='Password ...' onclick='value=""'/><p /> <input type='submit' name='submit' value='Login to my Account' /> </form> I would really appreciate it if someone could help me, thanks. P.S. If you need the php part of the header file it is here: <?php session_start(); include ( './includes/functions.php' ); include ( './includes/connect_to_mysql.php' ); ?>

    Read the article

  • mysql to codeigniter active record help

    - by JoeM05
    Active record is a neet concept but sometimes I find it difficult to get more complicated queries to work. I find this is at least one place the CI docs are lacking. Anyway, This is the sql I wrote. It returns the expected results of quests not yet completed by the user that are unlocked and within the users level requirements: SELECT writing_quests . * FROM `writing_quests` LEFT OUTER JOIN members_quests_completed ON members_quests_completed.quest_id = writing_quests.id LEFT OUTER JOIN members ON members.id = $user_id WHERE writing_quests.unlocked =1 AND writing_quests.level_required <= $userlevel AND members_quests_completed.user_id IS NULL This is the codeigniter active record query, it returns all quests that are unlocked and within the users level requirement: $this->db->select('writing_quests.*'); $this->db->from('writing_quests'); $this->db->join('members_quests_completed', 'members_quests_completed.quest_id = writing_quests.id', 'left outer'); $this->db->join('members', "members.id = $user_id", 'left outer'); $this->db->where('writing_quests.unlock', 1); $this->db->where('writing_quests.level_required <=', $userlevel); $this->db->where('members_quests_completed.user_id is null', null, true); I'm guessing there is something wrong with the way I am asking for Nulls. To be thorough, I figured I'd include everything.

    Read the article

  • How to check if a child-object is populated

    - by TheQ
    How can i check if a child-object of a linq-object is populated or not? Example code below. My model have two methods, one joins data, and the other does not: public static Member GetMemberWithPhoto(Guid memberId) { using (DataContext db = new DataContext()) { DataLoadOptions dataLoadOptions = new DataLoadOptions(); dataLoadOptions.LoadWith<Member>(x => x.UserPhoto); db.LoadOptions = dataLoadOptions; var query = from x in db.Members where x.MemberId == memberId select x; return query.FirstOrDefault(); } } public static Member GetMember(Guid memberId) { using (DataContext db = new DataContext()) { var query = from x in db.Members where x.MemberId == memberId select x; return query.FirstOrDefault(); } } Then my control have the following code: Member member1 = Member.GetMemberWithPhoto(memberId); Member member2 = Member.GetMember(memberId); Debug.WriteLine(member1.UserPhoto.ToString()); Debug.WriteLine(member2.UserPhoto.ToString()); The last line will generate a "Cannot access a disposed object" exception. I know that i can get rid of that exception just by not disposing the datacontext, but then the last line will generate a new query to the database, and i don't want that. What i would like is something like: Debug.WriteLine((member1.UserPhoto.IsPopulated()) ? member1.UserPhoto.ToString() : "none"); Debug.WriteLine((member2.UserPhoto.IsPopulated()) ? member2.UserPhoto.ToString() : "none"); Is it possible?

    Read the article

  • Can I lock tables in an IF statement in MySQL?

    - by MalcomTucker
    This is throwing a syntax error - --from body of a stored proc IF (name = in_name) SET out_id = temp; ELSE LOCK TABLE People WRITE; INSERT INTO People (Name) VALUES (in_name); UNLOCK TABLE; SELECT LAST_INSERT_ID() INTO out_id END IF do I have to lock any tables I need at the start of the SP?

    Read the article

  • How can i solve "Captcha required" error in Google Apps API Ver 2 for .NET ?

    - by Preeti
    Hi, I am migrating Contacts to Google Apps. But after migrating around 300 contacts I am getting "Captcha Required" Exception at line : Uri feedUri = new Uri(ContactsQuery.CreateContactsUri(UserName)); ContactEntry createdEntry = (ContactEntry)service.Insert(feedUri, ContactEntry[0]); I am using Ver2 of Google API. How can i solve this issue ? Note : I am not using web application. Thanx

    Read the article

  • Selecting More Than 1 Table in A Single Query

    - by Kamran
    I have 5 Tables in MS Access as Logs [Title, ID, Date, Author] Tape [Title, ID, Date, Author] Maps [Title, ID, Date, Author] VCDs [Title, ID, Date, Author] Book [Title, ID, Date, Author] I tried my level best through this code SELECT Logs.[Author], Tape.[Author], Maps.[Author], VCDs.[Author], Book.[Author] FROM Logs , Tape , Maps , VCDs, Book WHERE ((([Author] & " " & [Author] & " " & [Author] & " " & [Author]& " " & [Author]) Like "*" & [Type the Title or Any Part of the Title and Press Ok] & "*")); I want to select all of these fields in a single query. Suppose there is Adam as author of works in all tables. So when i put Adam in search box it should result from all tables. I know this can be done by having single table or renaming fields names but that's not required. Please help.

    Read the article

  • Modify php shopping cart to support multiple drop down menus

    - by Thomas
    I have a shopping cart script that I am trying to modify to support multiple product selection. As it is now, the customer can select a product from a single drop down menu. Now, I would like to add multiple dropdown menus (all populated with the same options). Here is the php that outputs the dropdown menu: if($eshopoptions['options_num']>1){ $opt=$eshopoptions['options_num']; $replace.="\n".'<label for="eopt'.$theid.'"><select id="eopt'.$theid.'" name="option">'; for($i=1;$i<=$opt;$i++){ $option=$eshop_product['products'][$i]['option']; $price=$eshop_product['products'][$i]['price']; if($option!=''){ if($price!='0.00') $replace.='<option value="'.$i.'">'.stripslashes(esc_attr($option)).' @ '.sprintf( _c('%1$s%2$s|1-currency symbol 2-amount','eshop'), $currsymbol, number_format($price,2)).'</option>'."\n"; else $replace.='<option value="'.$i.'">'.stripslashes(esc_attr($option)).'</option>'."\n"; } } Is there some really simple way of getting the code to output the menu say 3 times instead of once?

    Read the article

  • Trouble editing Word document with PHP

    - by bhoomi-nature
    I want to open a word document and edit it I am opening the word document from the server and at that time it's opening with garbage values (perhaps it's not being properly converted to UTF-8). When I delete those garbage values and insert something from a textarea to that file it is going to insert and from then on it opens properly. I would like the document to open with the English words in the document instead of garbage value - it's only the first opening that's broken at present. <? $filename = 'test.doc'; if(isset($_REQUEST['Submit'])){ $somecontent = stripslashes($_POST['somecontent']); // Let's make sure the file exists and is writable first. if (is_writable($filename)) { // In our example we're opening $filename in append mode. // The file pointer is at the bottom of the file hence // that's where $somecontent will go when we fwrite() it. if (!$handle = fopen($filename, 'w')) { echo "Cannot open file ($filename)"; exit; } // Write $somecontent to our opened fi<form action="" method="get"></form>le. if (fwrite($handle, $somecontent) === FALSE) { echo "Cannot write to file ($filename)"; exit; } echo "Success, wrote ($somecontent) to file ($filename) <a href=".$_SERVER['PHP_SELF']."> - Continue - "; fclose($handle); } else { echo "The file $filename is not writable"; } } else { // get contents of a file into a string $handle = fopen($filename, 'r'); $somecontent = fread($handle, filesize($filename)); ?> <h1>Edit file <? echo $filename ;?></h1> <form name="form1" method="post" action=""> <p> <textarea name="somecontent" cols="80" rows="10"><? echo $somecontent ;?></textarea> </p> <p> <input type="submit" name="Submit" value="Submit"> </p> </form> <? fclose($handle); } ?>

    Read the article

  • Persistance Queue Implementation

    - by Winter
    I was reading an article on Batch Processing in java over at JDJ http://java.sys-con.com/node/415321 . The article mentioned using a persistence queue as a Batch Updater instead of immediately sending an individual insert or update to the database. The author doesn't give a concrete example of this concept so I googled Persistence Queue but that didn't come up with much. Does anyone know of a good example of this?

    Read the article

  • jQuery ajax response not operating correctly

    - by mmarceau
    Ok this is frustrating... The code below works "correctly" as far as sending the email address to the SaveEmail URL and it gets saved correctly each time I change the drop down. However it only outputs the "Successful" message once, no matter how many times I change the value in the drop down. The "data" that is returned is "Successful". I would like to show the message for a couple seconds, then fade it out. It works correctly the first time I change the drop down, after that the change happens and the value gets saved, but the "Successful" message doesn't display. jQuery code: $('#AgentEmails').change(function() { var NewAddress = $('#AgentEmails').val(); $.post('SaveEmail.aspx', { email: NewAddress }, function(data) { $('#SelectMsg').html("<b>" + data + "</b>").fadeOut(); }); }); HTML code: <select ID='AgentEmails' runat='server'> <option value="[email protected]">TEST</option> </select><span id='SelectMsg'></span> What needs to be changed in my code to make this operate correctly? Thanks for the help.

    Read the article

  • Not able to add column value

    - by peter
    I have project which contain 4 tables among these shedule table Session column i am not able to add vaue,,this table contain three foriegn key from two tables(in which single table has two foreign keys here) i added values here..Any one has any idea about this..Actually my intention is to remove the error "the insert statement conflicted with the foreign key constraint sql server"

    Read the article

  • Further filter SQL results

    - by eric
    I've got a query that returns a proper result set, using SQL 2005. It is as follows: select case when convert(varchar(4),datepart(yyyy,bug.datecreated),101)+ ' Q' +convert(varchar(2),datepart(qq,bug.datecreated),101) = '1969 Q4' then '2009 Q2' else convert(varchar(4),datepart(yyyy,bug.datecreated),101)+ ' Q' +convert(varchar(2),datepart(qq,bug.datecreated),101) end as [Quarter], bugtypes.bugtypename, count(bug.bugid) as [Total] from bug left outer join bugtypes on bug.crntbugtypeid = bugtypes.bugtypeid and bug.projectid = bugtypes.projectid where (bug.projectid = 44 and bug.currentowner in (-1000000031,-1000000045) and bug.crntplatformid in (42,37,25,14)) or (bug.projectid = 44 and bug.currentowner in (select memberid from groupmembers where projectid = 44 and groupid in (87,88)) and bug.crntplatformid in (42,37,25,14)) group by case when convert(varchar(4),datepart(yyyy,bug.datecreated),101)+ ' Q' +convert(varchar(2),datepart(qq,bug.datecreated),101) = '1969 Q4' then '2009 Q2' else convert(varchar(4),datepart(yyyy,bug.datecreated),101)+ ' Q' +convert(varchar(2),datepart(qq,bug.datecreated),101) end, bugtypes.bugtypename order by 1,3 desc It produces a nicely grouped list of years and quarters, an associated descriptor, and a count of incidents in descending count order. What I'd like to do is further filter this so it shows only the 10 most submitted incidents per quarter. What I'm struggling with is how to take this result set and achieve that.

    Read the article

  • Beginner MVC question - Correct approach to render out a List and details?

    - by fizzer
    I'm trying to set up a page where I display a list of items and the details of the selected item. I have it working but wonder whether I have followed the correct approach. I'll use customers as an example I have set the aspx page to inherit from an IEnumerable of Customers. This seems to be the standard approach to display the list of items. For the Details I have added a Customer user control which inherits from customer. I think i'm on the right track so far but I was a bit confused as to where I should store the id of the customer whose details I intend to display. I wanted to make the id optional in the controller action so that the page could be hit using "/customers" or "customers/1" so I made the arg optional and stored the id in the ViewData like this: public ActionResult Customers(string id = "0") { Models.DBContext db = new Models.DBContext(); var cList = db.Customers.OrderByDescending(c => c.CustomerNumber); if (id == "0") { ViewData["CustomerNumber"] = cList.First().CustomerNumber.ToString(); } else { ViewData["CustomerNumber"] = id; } return View("Customers", cList); } I then rendered the User control using RenderPartial in the front end: <%var CustomerList = from x in Model where x.CustomerNumber == Convert.ToInt32(ViewData["CustomerNumber"]) select x; Customer c = (Customer)CustomerList.First(); %> <% Html.RenderPartial("Customer",c); %> Then I just have an actionLink on each listed item: <%: Html.ActionLink("Select", "Customers", new { id = item.CustomerNumber })% It all seems to work but as MVC is new to me I would just be interested in others thoughts on whether this is a good approach?

    Read the article

  • Connecting FLEX 3.0, JAVA WITH MYSQL

    - by Nithi
    How to connect to MYSQL DB from Java, create table, insert data, retrieve it with datatypes. How to make use of the data to/from in Flex application. plz help me out.. i have basic knowledge in sending and receiving messages using BlazeDS. and calling JAVA METHODS USING ...

    Read the article

  • How do you debug MySQL stored procedures?

    - by Cory House
    My current process for debugging stored procedures is very simple. I create a table called "debug" where I insert variable values from the stored procedure as it runs. This allows me to see the value of any variable at a given point in the script, but is this is there a better way to debug MySQL stored procedures?

    Read the article

< Previous Page | 545 546 547 548 549 550 551 552 553 554 555 556  | Next Page >