hello all Can anybody help me out with the Php and Mysql code. The mysql query does not fetch data from the database. The example file are mentioned below:
This is the code for making a drop down list. <html> <head>
<script> function showUser(str) { if (str == "") { document.getElementById("txtHint").innerHTML = ""; return; } else { 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 (this.readyState == 4 && this.status == 200) { document.getElementById("txtHint").innerHTML = this.responseText; } }; xmlhttp.open("GET","isolates.php?q="+str,true); xmlhttp.send(); } } </script></head>
<body>
<select name="q" onchange="showUser(this.value)">
<option value="">Select a strain:</option>
<option value="1">ABNIH5</option>
<option value="2">ABNIH5</option>
<option value="3">ABNIH5</option>
<option value="4">ABNIH5</option>
</select>
</form>
</body> </html>
and this is another code to fetch data. $q = $_GET['q']; $dbhost = 'abc'; $dbuser = 'abc'; $dbpass = 'abc'; $conn = mysql_connect($dbhost, $dbuser, $dbpass); if (! $conn) { die ('could not connect: ' . mysql_error()); }
if (!mysql_select_db('xyz', $conn)) { echo 'Could not select database'; exit; }
$sql= "SELECT Genome_id,Genome_name,Genome_status,Completion_date,Publication FROM tablename WHERE Strain_id LIKE '%".$q."%'";
$result = mysql_query($sql, $conn);
echo "
| Genome_id | Genome_name | Genome_status | Completion_date | Publication |
|---|---|---|---|---|
| " . $row['Genome_id'] . " | " . $row['Genome_name'] . " | " . $row['Genome_status'] . " | " . $row['Completion_date'] . " | " . $row['Publication'] . " |
?>
</body> </html>
