我想使用php中的AJAX方法更新状态。在php代码中,我添加了select命令以将状态更改为完成,并将值传递到update.php文件中。但没有改变发生
AJAX Code:
<script>
function updatestatus(status) {
if(str == '') {
document.getElementById("res").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 (this.readyState==4 && this.status==200) {
document.getElementById("res").innerHTML=this.responseText;
}
}
xmlhttp.open("GET","update.php?id=$row[id]"+str,true);
xmlhttp.send();
}
PHP Code : I need to update the status from assigned to complete, plz help me in this regard.
while ($row = mysqli_fetch_array($res)) {
echo "<tr>";
echo "<td>".$row['project'];"</td>";
echo "<td>".$row['date'];"</td>";
echo "<td>".$row['tl_name'];"</td>";
echo "<td>".$row['subject'];"</td>";
echo "<td>".$row['details'];"</td>";
echo "<td>
<form method='POST' action=''>
<select name='status' id='status'>
<option value='Assigned'>Assigned</option>
<option value='Completed'>Completed</option>
</select>
</form>
</td>";
echo "<td><input type='submit' id='button' name='button' onsubmit='updatestatus(this.value)' value='UPDATE'></td>";
echo "</tr>";
}
In the update.php am using the below code:
$status = $_POST['status'];
$id = $_GET['id'];
$sel = "update workassign set status ='$status' where id ='$id'";
$res = mysqli_query($conn,$sel);
我需要将状态从已分配更新为完成,请在这方面帮助我。