rilana Posted June 19, 2010 Share Posted June 19, 2010 Hi Guyes I didn't do php for a while and now I have to update a project I did a while back. So I dicided to throw some AJAX in there... So I grabed some tutorial and now I am kind of back to scratch. I would really aprechiate your help. I have a database, a html and a php file. Rightnow things looke like this: http://neu.smartpersonal.ch/stellen.php At first looke it seems to work, but then I realized that it puts jobs with the same ID out more then once. Much more then once... My HTML looks like that: <script type="text/javascript"> function showUser(str) { if (str=="") { document.getElementById("txtHint").innerHTML=""; return; } if (window.XMLHttpRequest) {// code for IE7+, Firefox, Chrome, Opera, Safari xmlhttp=new XMLHttpRequest(); } else {// code for IE6, IE5 xmlhttp=new ActiveXObject("Microsoft.XMLHTTP"); } xmlhttp.onreadystatechange=function() { if (xmlhttp.readyState==4 && xmlhttp.status==200) { document.getElementById("txtHint").innerHTML=xmlhttp.responseText; } } xmlhttp.open("GET","stellenResult.php?q="+str,true); xmlhttp.send(); } </script> </head> <body id="stellen"> <div id="container"> <div id="banner" ><h1> </h1> </div> <div id="left"> <a href="index.php"><img src="pic/logoLeftWeb.gif" border="0" /> </a></div> <div id="right"> </div> <!-- end menu right --> <div id="content"> <div id="nav"><? include("nav.html"); ?> </div> <h3><strong>Finden</strong> Sie Ihre <strong>Gelegenheit</strong></h3> <form> <select name="users" onchange="showUser(this.value)"> <option value="">Kagegorie wählen:</option> <option value="1" />Technische Berufe</option> <option value="3" />Bau</option> <option value="4" />Verkauf</option> <option value="2" />Kaufmännische Berufe</option> </select> </form> <div id="txtHint"><b>Person info will be listed here.</b></div> And my PHP looks like that: <? //java $q=$_GET["q"]; // Make a MySQL Connection $con = mysql_connect("", "", "") or die(mysql_error()); if (!$con) { die('Could not connect: ' . mysql_error()); } mysql_select_db("", $con); mysql_query("SET NAMES 'UTF8'"); $sql = "SELECT * FROM stelle JOIN stelle_region ON stelle_region.stelleID=stelle.stelleID JOIN region ON region.regionID=stelle_region.regionID JOIN stelle_beruf ON stelle_beruf.stelleID=stelle.stelleID JOIN beruf ON beruf.berufID=stelle_beruf.berufID WHERE stelle_beruf.berufID = '".$q."'"; echo $sql; $result = mysql_query($sql); $r = mysql_fetch_row($result); $numrows = $r[0]; $numrows = mysql_num_rows ($result); //print table echo "<table width='100%' border='0' cellpadding='0'>"; echo "<tr height='40px'><td>"; echo "Ihre Suche ergab $numrows Treffer"; echo "</tr></td><td></td></tr>"; while($row = mysql_fetch_array( $result )) { $id=($row['stelleID']); $datumsanzeige=($row['stelleDatumsanzeige']); $datum=(date("d.m.y", strtotime($row["stelleDatum"]))); $today = date("d.m.y"); if($datumsanzeige == "fake") { $datumDef = "$today"; } else { $datumDef = "$datum"; } // Print out the contents of each row into a table echo "<tr><td class='stellen'>"; echo $row['stellePosition']; echo "<br>"; echo $row['berufName']; echo "<br>"; echo $row['stelleID']; echo "</td><td>"; echo "$datumDef"; echo "</td></tr><tr><td>"; //take <li> away $text=strip_tags ($row['stelleStellenbeschrieb']); // 0,100 show 100 Zeichen echo substr($text,0,100); echo "... "; echo "<a href=\"stellenDetail.php?stelleID=$row[stelleID]\">|weiter > </a>"; echo "</td></tr>"; echo "<tr><td>"; echo "</td></tr>"; echo "<tr><td colspan='2' class='line'><br></td><td></td></tr>"; } echo "<tr><td>"; echo "</td></tr>"; echo "</table>"; echo mysql_error(); mysql_close($con); ?> I dont understand why it would put the same id's out more then once...??? And I dont see a pattern in it eighter. I would really aprechiate some help or hints thank you so much! Rilana Link to comment https://forums.phpfreaks.com/topic/205274-php-ajax-and-mysql-output-works-but-the-output-is-multiple-with-same-id/ Share on other sites More sharing options...
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