kbh43dz_u
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Posts posted by kbh43dz_u
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are your settings really saved? (do they still exist when you reopen the config file?) ...you should try restarting your webserver (software not hardware). If it's still OFF there must be another PHP.ini anywhere else.
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<div id="myDiv" style="display:none"></div>
and when you pass something bigger to your javascript function it could do this to make it appear:
document.getElementById("myDiv").style.display = "block";
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but this is still not php but javascript ... if you want php only there will be no way - except setting a timeout for the script (but the client will not get the result until the timeout is over)
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are you talking about javascript or java-server-pages (jsp)?
your javascript has to be in javascript-tags like <script>function myFunction(){}</script>
for more details we need your script.
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... I just got it running ... the module .load and .conf were missing and incorrect.
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...I would need it for debian etch! (but if it doesn't work i could change distribution)
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Hello!
I need PHP6 (snapshot) installed for a project and it is just not working >.> Did anyone try to install it? I used 4 tutorials but I couldn't figure out how to get it working. I already managed to have it compiled completely but then the mudule-files (.so) were missing and I couldn't integrate it.... there must have been a lot wrong with my way to install it.
Does anyone know more about, or have other tutorials.... or maybe someone could tell me how he installed it....
best regards!
(used tutorials:
http://blog.agoraproduction.com/index.php?/archives/19-Installing-PHP6-For-beginners.html
http://www.v-nessa.net/2007/02/26/how-to-install-php6
http://www.bencornwell.com/2007/11/18/php-6-installation-guide-for-ubuntu-710-gutsy-gibbon/
http://www.journaldunet.com/developpeur/tutoriel/php/071106-php6-apache-ubuntu.shtml
)
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Try
ALTER TABLE tablename AUTO_INCREMENT = 1
It should set auto increment to the number of your last entry + 1. (Setting it to 0 when you have entries will not work )
I hope I could help you.
edit: if you want to add an entry with an auto increment number which you deleted before you can use:
SET insert_id = 8; INSERT INTO tablename VALUES (’field1’, '…');
(for example if you have entries from 0 to 10 but deleted entry number 8 )
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change line1 and line2?
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Sure it can. But why?
You'll have to write your own hooks, since the standard php library will not contain methods to do so.
*g*
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$domain = "xyz.com";
$entry = dns_get_record($domain, DNS_MX);
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try " include" instead of "require"
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Hello, I am new to javascript and tried modifying an existing script but it doesn't work. It is used to see if one option on a form is selected, and the other form field is empty, it will error. Here is what I modified:
<script language="JavaScript" type="text/javascript"> <!-- function check_music() { if (form.music_fan.value == 1 && form.musician_name.value == "") { alert( "test." ); form.music_fan.focus(); return false ; } return true ; } //--> </script>
and on the first <form I added:
onSubmit="return checkme_signup()"
There are no errors but it just continues to the next page. I used the php && so perhaps javascript uses something else for "and" or something. Basically I only want it to alert when the music_fan value is = 1 and the musician_name is empty. If the music_fan value is 0 or the musician_name is not empty it will not error. Thanks!
..>
Your script (check music() should look like that:
<script language="JavaScript" type="text/javascript"> <!-- function check_music() { if (document.getElementById('music_fan').checked && document.getElementById('musician_name').value == "") { alert( "error - but my script works." ); document.getElementById('musician_name').focus(); return false; } return true; } //--> </script>
The Value of "Music_fan" is ALWAYS == 1 because you never change the value of it - you just select it or not (this is why it always alerts you). also if you select the other checkbox, the value of checkbox one doesn't change. You want to use ".checked" ... ask if it is checked. (you don't need the values of the checkboxes in your script.)
In my example I'm calling the fields with "document.getElementById('musician_name')" so you will have tho add ID's for your fields - it's better because than your calls are unique.
(I added id="music_fan" to checkbox 1; id="music_fan2" to checkbox 2 and id="musician_name" to the musician_name-field. If you also want to check for a selection in the drop-down field you have to extend this script)
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sorry, i missed that it's about Javascript
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simply use
echo $myVar[0];
you can access every letter from a string like an array.
edit:
if you still need it in an array:
for ($i = 0; $i < strlen($myVar); $i++){ $array[] = $myVar[$i]; }
or
for ($i = 0; $i < strlen($myVar); $i++){ $array[] = substr ($myVar, $i, 1); }
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Pretty cool login effects! first i thought it was AJAX
Best Regards
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this would make it harder but not impossible (just try - no security by obscurity) ... better you check if the user is allowed to view the message: your users probably have to login. than you have their user id's or names saved in a session. check if the user i allowed to view it - but by an id transported by the URL (take something a user cant manipulate).
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you probably want to use strip_tags($text, '<p><div>'); and strip all except allowed tags.
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Why not add in a box at the end that they write y or n for if the information is correct and when that is filled in it automatically is triggered
why no button "send" or "next"
->
is there a way if both values are good to automaticy post to the next page instead of making a button ?
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There will be hardly any better way, except if it happens often that you want to hide more div's at the same thime -> than you could put all wanted divs in an array and give the array to your function. But also then the function does everything again for every div. So it will not be a big deal
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function openWin(width, height, scroll){ //scroll = yes/no win = window.open("test.php?action=view_users", "Window", "width="+width+",height="+height+",scrollbars="+scroll); }
<a href="terms.php" onclick="openWin("600", "400", "yes");return false">Terms</a
(are you sure you don't want the url for the popup to be dynamic? in the "new" function you posted it will not be given to the function by your link)
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thank you im a little scared though cause i know nothing about java and have only been learning php for a couple of months in my spare time.
Never ever call javscript "java"! completely different thing
No you cant't submit a form with PHP (PHP runs on the server) but with javascript.
The way ILYAS415 told you will work! but you probably don't want the form to be submitted right after you loaded the page, so you'll have to tell the function "submit()" WHEN you want to submit, like:
<script lanugage='javascript'> function checkForm(){ if(...){ document.form1.submit(); } } window.setInterval(checkForm_ress,100); //check if form has to submit again and again. </script>
instead of
<script lanugage='javascript'>document.form1.submit();</script>
But your problem will be: How do you know if your user has finished his input? (because if it is a search-form you don't know what he enters). (this is why I couldn't fill the "if clause" in my example)
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don't think so - its not designed for it.i don't know functions which catch keystrokes from the server (server = client in your case)
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please change my username to "kbh43dz_u". thx a lot!