me1000

Here is my problem, I have created a div that has a width of 100% Within the div there are 2 images, the right and left "ends". problem: The left end looks fine, but the right end wraps to the next line (it is still aligned to the right though) in Firefox. It looks fin in Safari, and but I haven't tried it in IE... here is the HTML, Ill post the css if requested, however i dont think that will be required, because I dont think the CSS is the problem. <div class="space"><img src="images/left.gif" align="left;" height="19px" width="14px" /><img src="images/right.gif" align="right" height="19px" width="15px" /></div> Again the "space" class is 100% width, so it is wider than both of those images combined, Any ideas? Thanks,

ok well my page has a list of each letter my links looks similar to this, <a href="letter.php?letter=a>A</a> <a href="letter.php?letter=b>B</a> and so on, problem with that is iUI reformats the link (using javascript) so the variable is never passed through when the page is rendered on the server. is there anyway to pass a variable for php to read through the link, in a different way?

So I am using iUI to develop my iPhone application, problem is I need to be able to pass a variable through a URL without using a link such as page.php?var=123 iUI reformats the URL, and the variable is not saved as a result. so is there a way I can make a link setting a value to a variable, but have it saved by the browser or something, and not the url. thanks,

This is what i finally got to work, $(document).ready(setTestimonials); function setTestimonials() { $('#testialink').click( function(){ $('#testidrop').slideToggle('slow') } ); $('#close').click( function(){ $('#testidrop').slideUp('slow') } ); }

Thank you for your reply, $('testialink').click(function(){$(document.getElementById('testidrop')).slideToggle("slow")}); Is that what you mean? That didn't work...

Is anyone familiar with jQuery? I have a div called testidrop, <div class="testidrop" id="testidrop">...</div> normally is is hidden, but when I click on a link (testialink) it should drop down, <a href="#" id="testialink" class="testialink">Testimonials</a> So I thought it was all working great, until I discovered than whenever you click ANYWHERE on the page it drops down! Here is the code in the head tag <script> //show testimonials $(document.getElementById('testialink')).click(function(){$(document.getElementById('testidrop')).slideToggle("slow")}); $(document.getElementById('close')).click(function(){$(document.getElementById('testidrop')).slideup("slow")}); </script> (BTW the reason for the close one, us because there is a link within the testidrop div that will close it) any idea why this happens? or what I can do to fix it? Big Thanks,

alright, thanks! you just made my life a lot easier! Thanks again!

not exactly what I want, because the text will vary, I thought you were asking if i was only interested the [verb] [/verb] tags, so which way would be best? your way or Jewbilee's way?

at the moment just this one tag is all i am interested in, and I am all for simple,

oops sorry, I forgot to reread my post, $string = 'The dog [verb]jumped[/verb] very high'; but then only display the word 'jumped'

Does anyone know of a function to ignore everything that is not within 2 defined strings? for example, 'The dog [verb]jumped[/very] high' but only display the word 'jumped' thanks

ok I think i have it, $pages = explode(" ", $info); foreach ($pages as $v) { $i++; } echo $i-1;

That gives me back the number of entries with ' ' in it. I just need to find the number of times ' ' is found within one entry

Is there a PHP function that counts the number of time something appears in a MySQL query? So I run the query, and it finds the number of times " " is found in the returned results.

Thank You, I really do appreciate it!

Woohoo! 1 bug down, since I think making a new topic for this next question would be unnecessary, I am not getting this error when running the script, Unknown column 'A' in 'where clause' not sure why I get this, because it shouldnt be calling for a column 'A', it is calling for column 'LETTER' where the value is either 'A' or 'a'

So i tried it, and got this back "Showing rows 0 - 0 (1 total, Query took 0.0003 sec)" ok, added that, and I will keep that in mind for future referance, Really all I care about is making sure there is more than one entry, at this point, I need all the information queried later on I just left all that part out for simplicity sake. my apologies for not being clear, I meant that if the variable is not defined it comes back as "A" however it defines just fine as any other letter. Thanks for you advise, im guessing it is on the database end since it is not coming back from the mysql client. here is what the row looks like, maybe that is of some help. i am just clueless at this point,

it seems Im having a similar problem, however with a different part. Why is this so difficult? anyway, $inLetterID = (isset($_REQUEST['letter'])) ? $_REQUEST['letter'] : "A"; echo $inLetterID; This always prints "A" (unless otherwise defined) $sqlquery = "SELECT * FROM INFORMATION WHERE LETTER like $inLetterID"; $num_pages = mysql_num_rows($sqlquery); echo '<br>'.$num_pages.': NUMBER OF PAGES<br>'; (for simplicity sake, i did not include the loop) the above puts this on the screen, no number (the number should return "1") here are some DB details, Table name: INFORMATION Column name: LETTER - (VARCHAR; 1 character ) any ideas? Thanks,

Thank You two, That fixed it! there was also another problem, however it was minor, I think it was just the quotes. thanks again!

Hi, So I have a table, in it there is a bunch of images that have a field that contains a number for which they are associated with. here is a partial screenshot now I need to get the images from the page, but only if there are images. if there are no images associated with the page defined, then it will not display anything. $inPageID = (is_numeric($_REQUEST['page'])) ? $_REQUEST['page'] : "1"; $numpic = mysql_query("SELECT * FROM VISUALS WHERE ASSO=$inPageID"); $num_pic = mysql_num_rows($numpic); if ($num_pic > "0" ) { echo '<div>'; echo 'it found pics'; echo '</div>'; } trouble is it doesn't work! ??? however if I change the greater than sign if ($num_pic > "0" ) to an equal sign it doesn't work either. so just out of curiosity I changed it to a less than sign, and what do you know, it worked! So since I know it is not returning a negative number is it returning a null set? Can anyone help? Thanks,

DUDE! you are my HERO!!!!! Thanks a million!

hia, So I have a column in a database called ID which is set to auto-increment so when I add a new entry to the database the number is one larger than the previous entry. is there an easier way than this $query = "SELECT ID FROM table ORDER DESC LIMIT 1"; $page_id = mysql_fetch_array(mysql_query($query)); to get the value of ID in the last entry?

That is interesting, Thanks,

That works, but it isnt exactly what I want, I would like to keep it a php redirect if I can! Thanks

Can someone please help me?? ??? ??? ??? ??? if (isset($inType)) { $query = "INSERT INTO $sTableName (TYPE, LINK, DATE, DESCRIPTION) VALUES ('$inType', '$inLink', '$inDate', '$inDescription')"; $result = mysql_query($query); $response = ($result) ? "Record Added" : "Error adding record!"; echo $response; header ("Location: process.php"); } Everything works great it even comes back and says "Record Added" All data is entered into the Database just like it should be! BUT IT WONT REDIRECT! what am I missing? lol I have looked at it for about 30 minutes now, maybe some fresh eyes will help! Thank You,