Call javascript from php in while loop

Question

I hope someone can help me, I've been struggling, I have a from and within the form is table which has columns. It retrieves the data from a Mysql table. In column 2 it displays the current information and in column 3 it displays the form options to update the values, eg:

First name:  John  [Form field to update]
Date of birth: 1970-01-01 [form field to update]

I'm specifically struggling to get the java script functioning for the popup calendar. If I run it from html it works fine but when I do it within php I can't get it to run because it displays the form within a while loop:

<head>
   <script src="datetimepicker_css.js"></script>
</head>
<body>

<?php

.....

$sql2="SELECT * FROM em_detail WHERE id = '$data1'";
$result2=mysql_query($sql2);

echo "<form action='upd_emp.php' method=post>";

while($row2 = mysql_fetch_assoc($result2)) {
$id_2 = $row2['id'] ;
.....

echo "<tr>";
echo "<td height=35px><strong>Date of Birth: </strong></td>";
echo "<td>$dob_2</td>";
echo "<td><input size=30 type=Text name=dob id=demo1/></td>";
echo "<td align=left><img src=images/cal.gif onclick=javascript:NewCssCal('demo1') style=cursor:pointer/></td>";
echo "      </tr>";

What does it do? What does it say in the error log or console? PHP just outputs HTML, it doesn't interact with the Javascript calls at all. It looks like the HTML you are outputting is invalid. Most likely that's the source of your problem. — Cfreak
– Cfreak, Commented Jul 8, 2011 at 16:00
in keeping with Cfreak's comment: php emits fixed pages from the server so the client computer is completely unaware of what happens there. If something is broken, you need to troubleshoot the fixed page that the browser sees, then figure out why the php has written the html in that manner. I only mention this because your title about "calling javascript from php" seems to indicate some confusion about the process. — horatio
– horatio, Commented Jul 8, 2011 at 18:35

Nicole · Accepted Answer · 2011-07-08 16:06:48Z

It's not working because in every row you print out in the loop, the corresponding inputs have the same id — demo1.

IDs in HTML should be unique across all elements. This means that every calendar that is opens looks for an input with ID of demo1, and it always finds the same one.

You'll want to make sure the input in each row has a unique id. You can use the variable you have, $id_2, or you can increment a counter for each iteration of the loop and use that. You will then use that unique number in both the id of the <input> and the onclick for the calendar trigger.

<?php
//...
$counter = 0;
while($row2 = mysql_fetch_assoc($result2)) {
$id_2 = $row2['id'] ;
?>
    <tr>
    <td height="35"><strong>Date of Birth: </strong></td>
    <td><?php echo $dob_2; ?></td>
    <td><input size="30" type="text" name="dob_<?php echo $id_2; ?>" id="demo_<?php echo $counter; ?>"/></td>
    <td align="left"><img src="images/cal.gif" onclick="NewCssCal('demo_<?php echo $counter; ?>');" style="cursor:pointer"/></td>
    </tr>
<?php

    $counter++;
}

Also I've update the code with the following changes:

name of an input should also be unique.
You should use quotes around attribute values in HTML
No need to echo your HTML code if it doesn't have variables - just use ?> and <?php when you want to get back into PHP code.
javascript: is not needed in onclick attributes — they are always Javascript, and the javascript: just becomes a label that isn't used.

Ryan · Accepted Answer · 2011-07-08 16:02:15Z

0

I believe the problem is the code you're trying to execute is using the ID in the input. ID's are only supposed to be used once on the page or JavaScript doesn't know what field to work with.

You have 2 potential options, the first is to use a class to call the date picker, this is what I do personally and it works a charm. You may need to modify your code a bit for this to work however.

The second is to dynamically generate a different ID for each input field you're using.

I hope this helps you.

answered Jul 8, 2011 at 16:02

Ryan

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2 Comments

Wilest Over a year ago

Thank you all so much! I've truly learned a great deal from you and got the code working now. Stackoverflow is super!

Ryan Over a year ago

@Wilest I'm glad to hear it, could you please mark the best answer for this question? Take care :-)

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