如何使用AJAX / PHP编辑mySQL表中的行？

Question

I have a mySQL table with entries each having an input ID and input text; can somebody help me fix this PHP/AJAX code so that I can edit the input text of the entries.

PHP CODE:

$inputID = $_POST['inputID'];
$inputtxt = $_POST['inputtxt'];

$strSql = "UPDATE Contributions SET $inputID = '$inputID',       $inputtxt='inputtxt' WHERE id = '$id'"

AJAX:

<script>
    $( "#EDITsave" ).click(function() {
    console.log("console")
    var index = document.getElementById("editinputtext").value;   
  $.ajax({
    type: "POST",
    url: "edit.php",
    data: {inputID: id, inputtxt:index},
    dataType: "text",
    success:function( msg ) {
      alert( "Data Saved: " + msg );
    }
   });
   });
 </script>

Answer 1

**

• Here is the PHP code, hope this works

**

$inputID = $_POST['inputID'];
$inputtxt = $_POST['inputtxt'];
$id=$_POST['id'];

$mysqli = new mysqli("localhost", "my_user", "my_password", "world");

/* check connection */
if (mysqli_connect_errno()) {
echo "Connect failed: %s
", mysqli_connect_error();
exit();
}

/* Prepare statement */
$query = "UPDATE Contributions SET inputtxt=? WHERE id = ?";
$stmt = $mysqli->prepare($query);

$stmt->bind_param($inputtxt, $inputID);

/* Execute the statement */
try{ 
$stmt->execute();
}catch(Exception $e){
echo $e->getMessage();
return;
}

/* close statement */
$stmt->close();

echo 'Record updated';

mysqli_close($link);