Echoing If Function?
Posted on
16th Feb 2014 07:03 pm by
admin
A script I am using has If statements in the comments form to basically tell the form what to do. Currently the form works by opening in a pop up for those who want to read/write comments. I'm changing it so that the form displays on the same page, essentially removing the need for a pop up. I have,
Code: if ($_GET['comment']) {
echo "<SCRIPT LANGUAGE='JavaScript'>
Did you know?Explore Trending and Topic pages for more stories like this.
function smiles(which) {
document.form1.text1.value = document.form1.text1.value + which;
}
</SCRIPT>
<form name='form1' method='post' action='comments.php?load=comments&view=add'>
<input type='hidden' name='aid' value='".$_GET['aid']."'>
<input type='hidden' name='name' value='".$_COOKIE[$pre."username"]."'>
<table cellspacing="0" cellpadding="1" border="0" align="center">
<tr><td><b>Name</b></td><td>".$_COOKIE[$pre."username"]."</td></tr>
As a snippet of some of the code. I want to know how I can make the If statement form appear on the page without removing,
Code: if ($_GET['comment']) {
As it would break the script since 'comment' is needed for the form to proceed without errors. I've tried variations of echo on $_GET['comment'] but it didn't do anything.
Hope that made sense, any ideas?
No comments posted yet
Your Answer:
Login to answer
173
13
Other forums
Strange HTML Tag?
I recently noticed some odd HTML appear in some of the websites I host. Not all of them are run on a
change text color with a jQuery code
Hihow can I change the text in a asp:TextBox to a different color when I start typing using jQuery?I
Can't seem to capture a variable in a chained select
I'm *this* close to having a chained select running but for some reason it doesn't seem to be pickin
PHP login form help (Done Most of It)
Hi i am having a problem, when i try logging in it is always saying "Invalid Login" im not
Displaying a record from mysql in a simple swf file
Hi,
I have a mysql database containing information I would like to display in my swf.
Hotlinking Picasa as the image folder of a website
Hi there PHP freaks, I would like to create a private album in Picasa to use it as the image folder
listcube issue
Hi
I have a virtual cube ZREMOTE. I use Tcode "listcube" to retrieve contents. I use of the
If a record matches an existing record do nothing
So I don't know if I should put this here or in mysql, but what my script is for is for people to ad
Remove Rows From Database ad
Hi All,
I have this:
Code: [Select]<?php
session_start();
include('../com
defining website tags
Hi,
What would be the best way to define tags for my site, such as website title, url etc.