uisneach Posted April 7, 2009 Share Posted April 7, 2009 Hello. I got this $sql = "SELECT * FROM notturna ORDER BY descrizione"; $result = mysql_query ($sql); while ($row = mysql_fetch_row($result)) { if ( $item == 0 ) { $_primafoto = $row[0]; } echo "photos[".$item."]=\"notturna/".$row[0].".jpg\";"; $item=$item+1; } } ?> function arrow() { document.getElementById( "back2" ).style.display = "none"; } function changePic(dir) { var image = document.images.photoslider, fwdBtn = document.getElementById('forward2'), backBtn = document.getElementById('back2'), n = photos.length-1; if (dir == "next") { which = (which < n) ? which + 1 : which; image.src = photos[which]; backBtn.style.display = "inline"; if (which == n) { fwdBtn.style.display = "none"; } } else if (dir == "back") { which = (which > 0) ? which - 1 : which; image.src = photos[which]; fwdBtn.style.display = "inline"; if (which === 0) { backBtn.style.display = "none"; } } return false; } </script> </head> <body OnLoad="arrow()"> <div class="container2"> <div id="logo" > <img title="logo" src="logo2.jpg"></div> <div class="menu"> <a href="http://paolobergomifoto.altervista.org">Home </a> <a href="http://paolobergomifoto.altervista.org">Chi sono </a> <a href="http://paolobergomifoto.altervista.org/gallerie.html">Gallerie </a> <a href="http://paolobergomifoto.altervista.org/contatti.html">Contatti </a> <a href="http://paolobergomifoto.altervista.org">Credits </a> </div> <div id="backnotturna"> <a href="#" onclick="return changePic('back');"> <img id="back2" style="border:0px" src="indietro.jpg"></a> </div> <div class="centro"> <div class="gruppofoto2"><a href="gallerymacro.php">Macro</a><a href="gallerypaesaggi.php">Paesaggi</a><a href="galleryritratti.php">Ritratti</a><a href="gallerybn.php">B&N</a><a href="gallerynotturna.php">Notturna</a><a href="galleryvarie.php">Varie</a><img src="notturna/<?php echo $_primafoto; ?>" name="photoslider"> </div> </div> <div id="forward"><a href="#" onclick="return changePic('next');"> <img id="forward2" style="border:0px" src="avanti.jpg"></a> </div> <div class="inizio"><a href="#" onclick="which=1; changePic('back');return false" >Torna all'inizio della gallery</a> </div> <div id="footer">Created by Paolo Bergomi</div> </div> The table in db has also a second field that in the query is represented by the variable $row[1] . this is the description of the pictures. I have an issue cause i don't know how to do to show the content of this field using the query above. The pictures is correctrly showed in the gallery, and my target is to add the relative description.(as i wrote, is the second field in the table) any idea ? cheers paolo Link to comment https://forums.phpfreaks.com/topic/153017-query-issue-print-the-second-field-of-a-my-sql-table/ Share on other sites More sharing options...
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