raman Posted November 2, 2008 Share Posted November 2, 2008 I want to show in HTML page what a particular image stands for by making that yellow box appear when mouse is pointed on that.Can someone tell how to do this ?I have already tried the alt attribute for the image but it does not work out,rather creates more problems. One of the example pages is : http://www.ncbs.res.in/index.php?option=com_content&task=view&id=349&Itemid=330 When you point on those three images on the right you can see their values "PDF" , "E-mail" etc. Link to comment https://forums.phpfreaks.com/topic/131061-show-the-value-of-a-image-html/ Share on other sites More sharing options...
dezkit Posted November 2, 2008 Share Posted November 2, 2008 is this what you are talking about Link to comment https://forums.phpfreaks.com/topic/131061-show-the-value-of-a-image-html/#findComment-680458 Share on other sites More sharing options...
raman Posted November 2, 2008 Author Share Posted November 2, 2008 No this is only a hyperlink. What I talked about I get with the attribute title.However I get an error,that is - instead of taking the value from the excel variable it goes into the second part of the loop,the else part. This is my code where I use the title attribute: echo"<p>The total number of records retrieved by your query is $counter.</p><input type='image' src='Excel.gif' name='excel' title='Download selected items as excel file.' /><input type='image' src='text.jpg' name='fasta' title='Download as fasta file' />"; And this is where I get the input from it: <?php $kyu=$_POST['pro']; $exc=$_POST['excel']; $fas=$_POST['fasta']; $conn= mysql_connect("localhost.localdomain","root","pichii13"); if(!$conn) { die('Could not connect:'.mysql_error()); } mysql_select_db("Protvirdb",$conn); if(empty($exc)) { $file='fastafile.fasta';$type="plain/text"; ob_start(); // echo"<p>"; foreach($_POST['pro'] as $kyu){ $re=mysql_query("SELECT * FROM fasta WHERE SNo='$kyu'"); while($row=mysql_fetch_array($re)) { echo"".$row['header']."";echo"\n";echo"".$row['Sequence']."";echo"\n"; } } }else{ $type="application/vnd.ms-excel;charset:UTF-8"; $file = 'ExcelFile.xls'; // start buffring ob_start(); // sample dynamically generated data echo '<table border="1"> '; echo '<tr><th>Category</th><th>Protein</th><th>Brief Description</th></tr>'; foreach($_POST['pro'] as $kyu){ $re=mysql_query("SELECT * FROM cryptovir WHERE SNo='$kyu'"); while($row=mysql_fetch_array($re)) { echo"<tr>"; echo"<td>".$row['Category']."</td>"; echo"<td>".$row['Name']."</td>"; echo"<td>".$row['Brief_Description']."</td>"; echo "</tr>"; } } } mysql_close($conn); $content = ob_get_contents(); ob_end_clean(); header("Expires: 0"); header("Last-Modified: " . gmdate("D, d M Y H:i:s") . " GMT"); header("Cache-Control: no-store, no-cache, must-revalidate"); header("Cache-Control: post-check=0, pre-check=0", false); header("Pragma: no-cache"); header("Content-type:$type;charset:UTF-8"); header('Content-length: '.strlen($content)); header('Content-disposition: attachment; filename='.basename($file));// // output all contents echo $content; exit; // If any tags/things not supported by excel will output then it will try to //open in office word ?> Link to comment https://forums.phpfreaks.com/topic/131061-show-the-value-of-a-image-html/#findComment-680471 Share on other sites More sharing options...
raman Posted November 2, 2008 Author Share Posted November 2, 2008 The problem is with giving two <input type=image" title='download' src="bla bla.jpg"><input type=image" title='dowddghd' src="blhdhdbla.jpg"> buttons in succession, IS THERE ANY WAY TO SORT THIS OUT? Link to comment https://forums.phpfreaks.com/topic/131061-show-the-value-of-a-image-html/#findComment-680514 Share on other sites More sharing options...
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