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08.05.2008 at 09:35AM PDT, ID: 23622849
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7.4

My db won't update with new data in PHP code

Asked by justmelat in PHP and Databases, MySQL Server

Tags:

i am using the attached code.  it just spits out 4 form fields that pull data from db and will update the info if the data is changed in the input boxes.  The first three boxes work just fine.  The fourth box will not update at all unless I go to the db and change the value there.  the value will be pulled into the input box, but if i change the value and click the "change", nothing happens, it will not update the value in that field.Start Free Trial
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Why won't db update with new data in PHP code
 
i am using the attached code.  it just spits out 4 form fields that pull data from db and will update the info if the data is changed in the input boxes.  The first three boxes work just fine.  The fourth box will not update at all unless I go to the db and change the value there.  the value will be pulled into the input box, but if i change the value and click the "change", nothing happens, it will not update the value in that field.
 
<fieldset>
<legend>Update From Emails</legend>
<table class="intake" border="0">
<?
$prjSubmittedTo=getSelectValues('Q_125');
?>
<tr>
 
<td width="25%"><b>Project Submitted to</b>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;</td>
<td width="75%"><b>From Email</b></td>
</tr>
 
<?php
$prjSubmittedTo;
foreach ($prjSubmittedTo as $key=>$value)
{?>
<tr>
<td><?=$value?></td>
<td><input type="text" maxlength="50" size="50" name="<?=Field($value)?>" value="<?=getcurrentfromemails(Field($value))?>"></td>
 
</tr> 
<?
}
?>
<tr>
<td>&nbsp;</td>
<td>
<input type="button" onclick="goTo('OPTIONS','emails');" value="Change">
</td>
</tr>
</table>
</fieldset>
 
=======================================
function updatefromemails($fromemails)
{
global $db;
$prjSubmittedTo = array();
 
$SubmittedTo=getSelectValues('Q_125');
foreach($SubmittedTo as $key=>$values)
{
array_push($prjSubmittedTo,Field($values));
}
 
foreach($fromemails as $key=>$values)
{	
if(in_array($key,$prjSubmittedTo))
{
$query="select * from email where PROJECT_SUBMITTED_TO='$key'";
$result=mysql_query($query);
$numrows=@mysql_num_rows($result);
if($numrows>0) 
{                                                                                                                 
$sql  = "UPDATE email SET email='$values' where PROJECT_SUBMITTED_TO='$key'";
}
else
{
   $sql ="INSERT INTO email (PROJECT_SUBMITTED_TO,email) VALUES ('$key','$values')";
}
 
mysql_query($sql,$db) or die("ERROR: " . mysql_error());
}
}
}
[+][-]08.05.2008 at 10:19AM PDT, ID: 22163029

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[+][-]08.05.2008 at 10:23AM PDT, ID: 22163068

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[+][-]08.05.2008 at 10:29AM PDT, ID: 22163117

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[+][-]08.05.2008 at 10:37AM PDT, ID: 22163198

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[+][-]08.05.2008 at 11:00AM PDT, ID: 22163357

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[+][-]08.05.2008 at 11:21AM PDT, ID: 22163528

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About this solution

Zones: PHP and Databases, MySQL Server
Tags: PHP, HTML, javascript, mysql
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Solution Provided By: justmelat
Participating Experts: 1
Solution Grade: A
 
 
 
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