I am trying to retrive value from database in dropdown and to display the datas related to the selected option.
I am new to php, Please Help me to write php code for the above .
My question 0 Light Poster
diafol
Ok, show us what you have and we'll guide you through it.
My question 0 Light Poster
index.php
<html>
<head>
<script>
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","getuser.php?q="+str,true);
xmlhttp.send();
}
</script>
</head>
<body>
<form>
<select name="users" onchange="showUser(this.value)">
<option value="">Select a person:</option>
<option value="1">Peter Griffin</option>
<option value="2">Lois Griffin</option>
<option value="3">Joseph Swanson</option>
<option value="4">Glenn Quagmire</option>
</select>
</form>
<br>
<div id="txtHint"><b>Person info will be listed here.</b></div>
</body>
</html>
getuser.php
<?php
$q = intval($_GET['q']);
$con = mysqli_connect('localhost','','','my_db');
if (!$con) {
die('Could not connect: ' . mysqli_error($con));
}
mysqli_select_db($con,"ajax_demo");
$sql="SELECT * FROM user WHERE id = '".$q."'";
$result = mysqli_query($con,$sql);
echo "<table border='1'>
<tr>
<th>Firstname</th>
<th>Lastname</th>
<th>Age</th>
<th>Hometown</th>
<th>Job</th>
</tr>";
while($row = mysqli_fetch_array($result)) {
echo "<tr>";
echo "<td>" . $row['FirstName'] . "</td>";
echo "<td>" . $row['LastName'] . "</td>";
echo "<td>" . $row['Age'] . "</td>";
echo "<td>" . $row['Hometown'] . "</td>";
echo "<td>" . $row['Job'] . "</td>";
echo "</tr>";
}
echo "</table>";
mysqli_close($con);
?>
The above code retrives the value from database properly, but i am trying to change the option value in index.php as
while($row = mysql_fetch_assoc($q))
{
$select .= "<option value='".$row['FirstName']."'>".$row['FirstName']."> </option>";
}
this is not working. please help me. thanks in advance
diafol
Am i right in thinking that this is nothing to do with ajax, you just need help creating the select field on page load filling it with data from your DB?
BTW - will FirstName be appropriate as a filter? Isn't is possible that many users will have the same first name?
My question 0 Light Poster
since i'm not getting it, can u plz guide me
diafol
I have no idea what you're trying to do.
My question 0 Light Poster
I'm trying to fetch value from database into dropdown, and if a option is selected then it displays the related record in table format for selected option in dropdown
Plz guide me to do this
Thanks in advance
diafol
This is an example - you need to change fieldnames, tablenames to suit yourself. In addition, I use PDO instead of mysqli (I don't like mysqli) - so you may need to modify that aspect slightly too.
$options = "<option value=''>Select a person:</option>\n";
$sql = "SELECT id, firstname, lastname FROM users";
foreach ($db->query($sql) as $row) {
$options .= "<option value='{$row['id']}'>{$row['firstname']} {$row['lastname']}</option>\n";
}
...
<select id="users" name="users">
<?php echo $options;?>
</select>
<div id="results"></div>
<script>
var sel = document.getElementById('users');
sel.onchange=function(){
//run your ajax code and update the html table in #results
//with data from the DB
};
</script>
My question 0 Light Poster
Thank you very much.Fianlly i got it
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