Jonas_R Posted September 23, 2006 Share Posted September 23, 2006 Hi,Iam a bit unsure where to put this topic, if it wrong i hope the admin's will move it.My problem is that i want to use a javascript inside a php code, and i cant seem to get it to work. Well it looks like this:Iam pretty sure my problem is, that i cant seem to make the correct use of " " and ' ' . [code]<script language="JavaScript" src="mitjs.js"></script> //here is a function called "na_open_window"</head><body><?phpinclude('test3.php');$query = mysql_query("SELECT id,path,parent_album FROM plogger_picturesWHERE parent_album = 3 ORDER BY id") or die(mysql_error());if (mysql_num_rows($query) == 0) { echo 'No pictures!<br>';} else { echo '<table width="800" border="0"><tr>'; echo'<td width="800">'; while($row = mysql_fetch_assoc($query)) { $file =$row[id]."-".basename($row[path]); $name = "Anne".$row[id]; //echo "<p>$file<p>"; echo "<a href='javascript:na_open_window('$name','ploggerb2.1/images/$row[path]', 0, 0, 0, 0, 0, 0, 0, 0, 0)'target='_self'><img src='ploggerb2.1/thumbs/$file' border='0'></a>"; echo " "; } echo'</td>'; echo'</tr>'; echo'</table>';}?>[/code]Best RegardsJonas Rohde Quote Link to comment https://forums.phpfreaks.com/topic/21784-problem-with-javascript-in-php-photoalbum/ Share on other sites More sharing options...
yonta Posted September 23, 2006 Share Posted September 23, 2006 It would help if you tell what it does output as it stands. But here are some suggestions.Normaly when getting values from a mysql query i do this$file =$row['id']."-".basename($row['path']); - notice the '' around column names. i'm not sure if they are mandatory but that's the commom way.Then here i would o this:echo '<a href="javascript:na_open_window(\''.$name.'\',"ploggerb2.1/images/'.$row['path'].'", 0, 0, 0, 0, 0, 0, 0, 0, 0)"target="_self"><img src="ploggerb2.1/thumbs/'.$file.'" border="0"></a>';Using ' instead of " is quicker because php doesn't execute strings inside single quotes - this also means that when you want to print a variable you have to break the string and use the concatenation operator (.). Also when you want to escape quotes (be it single or double) use the backslash \ (notice the \ before the single quotes in $name variable).See more at [url=http://www.php.net/manual/en/language.types.string.php]http://www.php.net/manual/en/language.types.string.php[/url] on handling strings.Hope this helps :) Quote Link to comment https://forums.phpfreaks.com/topic/21784-problem-with-javascript-in-php-photoalbum/#findComment-97380 Share on other sites More sharing options...
Jonas_R Posted September 23, 2006 Author Share Posted September 23, 2006 Hi,Thanks for the help :)It was, what i was looking for. Now everything works as it is supposed to.Thx very much./Jonas R. Quote Link to comment https://forums.phpfreaks.com/topic/21784-problem-with-javascript-in-php-photoalbum/#findComment-97400 Share on other sites More sharing options...
Recommended Posts
Join the conversation
You can post now and register later. If you have an account, sign in now to post with your account.