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PHP - How/what Script Can I Execute (on Page/form Load) In Order To Show/hide A Container?
Hi,
In reference to my first attached image, I have a form which displays two SELECT/drop-down fields (labeled "Store Name" and "Item Description".....and both of which pull-in values from two separate lookup/master tables, in addition to providing an additional option each for "NEW STORE" and "NEW ITEM").
Now, when first-run, and/or if "NEW STORE" and "NEW ITEM" are not selected from the drop-down's then the two fields in green ("New Store Name" and "New Item Name" are hidden, by means of the following code:
<div class="new-store-container" id="new-store-container" name="new-store-container" style="display:none;">
<div class="control-group">
<div class="other-store" id="new_store_name">
<?php echo standardInputField('New Store Name', 'new_store_name', '', $errors); ?>
</div>
</div>
</div>
Conversely, if "NEW STORE" and/or "NEW ITEM" are selected from the two drop-down's then one (or both) of the "New Name" fields are unhidden by means of the following two pieces of code, one PHP and the second JS:
<select class="store-name" name="store_id" id="store_id" onclick="toggle_visibility('store_id','new-store-container')">
<?php echo $store_options; ?>
<?php
if($values['store_id'] == "OTH")
{
echo "<option value='OTH' selected> <<<--- NEW STORE --->>> </option>";
}
else
{
echo '<OPTION VALUE="OTH"> <<<--- NEW STORE --->>> </OPTION>';
}
?>
</select>
function toggle_visibility(fieldName, containerName)
{
var e = document.getElementById(fieldName);
var g = document.getElementById(containerName);
if (e.value == 'OTH')
{
if(g.style.display == 'none')
g.style.display = 'block';
else
g.style.display = 'none';
}
}
All of that is working just fine. The problem I'm having is that when I click the "Create" button, after having left any one of the form fields blank, the two "New Name" fields are hidden again, which I don't want to happen i.e. I want them to remain visible (since the values of "store_id" and/or "item_id" are "OTH"), so that the user can enter values into one or both of them, without havng to click on the drop-down a second time in order to execute the "on-click" code.
The second attached image shows how the fields are hidden, after clicking "Create".
How can I achieve that? It would be greate if someone could cobble-up the required code and provide it to me, since I'm relatively new to this.
Thanks much.
Snap1.png 26.14KB
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Snap2.png 149.47KB
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Thanks! Hi there, Im working on my little project and I would appreciate your help. I have only basic knowledge of php, mostly I just copy some scripts that could be useful for me. Im trying to find some simple script that allows me to see the written text on web page no. 1 on webpage no. 2.. Something like send the form to email, except I dont want to send it on email, but different webpage. Something like different way of eshop, where you get your order shipped to email, but I wanna send this information to webpage. Is there such a script like Im describing? Probably is but I dont know how to search for this..
Thank You
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I have a cron job that executes this script every 2 minutes:
<?php
// LOAD WP-LOAD.PHP
require('/opt/bitnami/apps/wordpress/htdocs/wp-load.php');
// INCLUDE AND EXECUTE SCHEDULER.PHP
include('/opt/bitnami/apps/wordpress/htdocs/wp-content/themes/yeelloe/scheduler.php');
?>
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I've been stuck trying to execute a script for a few weeks now and I really need help.
The script is supposed to schedule social media posts via functions.php and wp-load.php but I get errors.
Can someone do me a favour?
Drop your email and I'll send you the details to connect to my database.
Thanks
Hi guys, I am new here, and I am a bit stuck with doing something unusual. I want to create a script that can turn on a program (dynamips and dynagen). So far, I have tried 'exec' and 'shell_exec'. Soon I realized that apache runs the commands as 'www-data' user (apache2 in ubuntu) and it's very limited on what you can execute. Is there any way to do that at all? What would be the best practice? I am not concerned about security as this is not a production environment... Thanks Hi all I have this problem on a server using php5, unix based, safe_mode is On globally, i have turned it off locally through php.ini. Ok, this is testing example script i used: $cmd = ( "php -v" ); $out = shell_exec( $cmd ); print $out; On my own server this returns php version. On this mentioned server i'm using (commercial) this causes complete server breakdown, when logged in with SSH, i can't even issue "ls" command after that, nor find and kill the process. What could be so wrong with it? I don't think calling php-cli would make any difference. Hi,
I am a newbie at php and I recently tried making a small php configuration that runs on the localhost. execution.php
<?php
echo "first script has been executed";
exec('execution2.php');
?>
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Any help would be appreciated, thanks! |
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