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| Programming PHP, Perl, Ruby on Rails, AJAX, HTML, XHTML, CSS, JavaScript, MySQL and any other coding topics. |
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| Account Suspended | Function breaks out of the layout? Hi, I have a function that echo's the navigation for a website, it runs perfectly, BUT, it breaks out of the layout. If i do; PHP Code: The function is; PHP Code: ![]() So if i have; PHP Code: PHP Code: |
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| NamePros Member | PHP Code: |
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| Account Suspended | Quote:
Also, i don't want it to echo the "home" one | |
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| | #5 (permalink) |
| NamePros Member | It doesn't matter what "title" is, if you aren't selecting it in your mySQL query, you can't call $row['title']. Try Code: $NAVresult = mysql_query("SELECT title FROM pages WHERE site='$domain' ORDER BY id");
Bruce |
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| | #7 (permalink) |
| NamePros Member | Alas, it was only a quick 2 seconds code whilst on the phone and after being awake for way too long. As you'll know there are tonnes of mistakes in the whole thing. Anyways, this may be of more help.. you'll still need to connect up to your db etc and everything else you need to do lol. PHP Code: |
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| | #8 (permalink) |
| NPQ's PA, Slave, and On Call Coder Technical Services | My contribution ![]() Instead of: PHP Code: PHP Code:
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