如何检查请求是否成功

Question

i am using contenteditable property in p tag .. the code is

<p contenteditable="true" id="Option1_<?php echo $i ?>" style="width:98%;border:4px thin black; background-color:#D6D6D6;font-size:18px;color:black;padding:3px "><?php echo '&nbsp;'.'A.'.'&nbsp;&nbsp;&nbsp'.$question1['Option1'];?></p> 
  <p  contenteditable="true" id="Option2_<?php echo $i ?>" style="width:98%;border:4px thin black; background-color:#D6D6D6;font-size:18px;color:black;padding:3px "><?php echo '&nbsp;'.'B.'.'&nbsp;&nbsp;&nbsp'.$question1['Option2'];?></P> 

and jquery to make a request to make request document).ready(function(){

$("p[contenteditable=true]").blur(function(){
       var msg = $(".alert");
       var newvalue = $(this).text();
       var field = $(this).attr("id");
       $.post("ajax.php",field+"="+newvalue,function(d){
           var data = JSON.parse(d);
           msg.removeClass("hide");
            if(data.status == '200'){
                msg.addClass("alert-success").removeClass("alert-danger");
            }else{
                msg.addClass("alert-danger").removeClass("alert-success");
            }
           msg.text(data.response);
           setTimeout(function(){msg.addClass("hide");},3000);//It will add hide class after 3 seconds
       });
   });
});

and then php to update my mysql database on receiving the request

    <?php
$response = NULL;
$status = http_response_code(406);
if(!empty($_POST)){
session_start();
 $mock_test_name=$_SESSION['mock_test_name'];
$num_of_sections = $_SESSION['num_of_sections'];
$school_name = $_SESSION['school_name'];
$class_name = $_SESSION['class_name'];
$section_name = $_SESSION['section_name'];
$con = mysqli_connect("localhost","root","","onlinetest");
      if (!$con)
      {
      die('Could not connect: ' . mysqli_error());
      }
      $table_space = "$school_name $class_name $section_name $mock_test_name";
      $table = str_replace(" ", "_", $table_space);
      $table_space1 = "$school_name $class_name $section_name";
      $table1 = str_replace(" ", "_", $table_space1);
      $table_space2 = "$table1 $table";
      $table2 = str_replace(" ", "_", $table_space2); 
      $table2 = strtolower($table2);
    foreach($_POST as $key=>$value){
         $key = strip_tags(trim($key));
        $value = strip_tags(trim($value));
        $explode = explode("_",$key);
        $user_id = $explode[1];
        $field_name = $explode[0];
        if(isset($user_id)){
            $update = false;
            $selectData = mysqli_query($con,"SELECT " + $field_name + " FROM " + $table2 + " WHERE question_id='" + $user_id + "'"); //Selecting data from MySql
            $result = mysqli_fetch_assoc($selectData); //Fetching Data
            if($result[$field_name]!==$value){ //Checking if the Value is modified
                $update = mysqli_query($con,"UPDATE" + $table2+ "SET" + $field_name+"="+$value+ "WHERE question_id='"+$user_id+"'"); //Updating MySQL if value is Modifie
            }
            //Update the users Table
            if($update){
                $response = "User Details Updated";
                http_response_code(200); //Setting HTTP Code to 200 i.e OK
            }else{
                $response = "Not Modified";
                http_response_code(304); //Setting HTTP Code to 304 i.e Not Modified
            }
        }else{
            $response = "Not Acceptable";
        }
    }
}
echo json_encode(array(
    "status"=>$status,
    "response"=>$response
));
?>

But i think the request is not made properly as the database is not getting updated.. Please tell me how to check if a request has been made... or am i making error somewhere in writing code ??

Answer 1

Your mysqli_query function is formed incorrectly. You have to escape out of the parentheses in order to drop in variables.

You need to do something like

mysqli_connect($con, "SELECT " + var1 + " FROM " + var2);

or you'll end up making a query for

EDIT: For a more apt example, the line

$selectData = mysqli_query($con,"SELECT $field_name FROM $table2 WHERE question_id='$user_id'");

should be

$selectData = mysqli_query($con,"SELECT " + $field_name + " FROM " + $table2 + " WHERE question_id='" + $user_id + "'");

You'll notice the main difference, especially in coloring between the two, signifying that in the first one, the variables $field_name, $table2, and $user_id are all being interpreted as part of the query. You don't want the NAME of the variable, you want the VALUE of it instead, so you need to concatenate the strings together.

This is just one of the multiple similar fixes you'll need to do for your multiple queries. Every place the editor is marking the thing you're trying to use as a variable as part of the string, take the same steps to concat the strings.