Php 将mysql结果检索到jquery

我的问题是我无法通过ajax检索mysql结果，请帮助

ajax代码：

 $.ajax({
    type: "POST",
    url: "do_find_courses.php",
    //data:{question_id:question_id,answer:answer},
    data:{user_id:user_id}, dataType:'json',
    success:function(msg) {
        alert ('asdasd')
        // $("#quiz_form,#demo1").addClass("hide");
        // $('#result').show();
        $('p').html(msg);
    }
 });

 $.ajax({
    type: "POST",
    url: "do_find_courses.php",
    //data:{question_id:question_id,answer:answer},
    data:{user_id:user_id}, dataType:'json',
    success:function(msg) {
        alert ('asdasd');
                    console.log(msg);//You should check output of this in browser console
        // $("#quiz_form,#demo1").addClass("hide");
        // $('#result').show();
        $('p').html(msg.cname);
    }
 });

PHP代码：

 $final=array();
 $sql_courses=mysql_query("SELECT course_id, course_name FROM course") or die (mysql_error());
 $row_courses = mysql_fetch_array($sql_courses);            
 $result=$row_courses['course_name'];
 //array_push($final,$result);
 //print_r($result);
 echo json_encode($result);

 $final=array();
 $sql_courses=mysql_query("SELECT course_id, course_name FROM course") or die (mysql_error());
 $row_courses = mysql_fetch_array($sql_courses);            
 $result=$row_courses['course_name']; // this will have first_coures_name (an string)
 $final['cname']=$result;
 //print_r($result);
 echo json_encode($final); //the output should be this {'cname':'first_course_name'}

---

更改PHP代码如下

$final = array();
$sql_courses = mysql_query("SELECT course_id, course_name FROM course") or die(mysql_error());
$row_courses = mysql_fetch_array($sql_courses);

echo json_encode($row_courses);

更改php代码如下：

 $.ajax({
     type: "POST",
     url: "do_find_courses.php",
     //data:{question_id:question_id,answer:answer},
     data: {
         user_id: user_id
     },
     dataType: 'json',
     success: function (msg) {
         $('p').html(msg.course_name);
     }
 });

---

发送时最好使用如下键值：

最好使用

console.log（变量）检查变量的内容

ajax代码：

 $.ajax({
    type: "POST",
    url: "do_find_courses.php",
    //data:{question_id:question_id,answer:answer},
    data:{user_id:user_id}, dataType:'json',
    success:function(msg) {
        alert ('asdasd')
        // $("#quiz_form,#demo1").addClass("hide");
        // $('#result').show();
        $('p').html(msg);
    }
 });

 $.ajax({
    type: "POST",
    url: "do_find_courses.php",
    //data:{question_id:question_id,answer:answer},
    data:{user_id:user_id}, dataType:'json',
    success:function(msg) {
        alert ('asdasd');
                    console.log(msg);//You should check output of this in browser console
        // $("#quiz_form,#demo1").addClass("hide");
        // $('#result').show();
        $('p').html(msg.cname);
    }
 });

PHP代码：

 $final=array();
 $sql_courses=mysql_query("SELECT course_id, course_name FROM course") or die (mysql_error());
 $row_courses = mysql_fetch_array($sql_courses);            
 $result=$row_courses['course_name'];
 //array_push($final,$result);
 //print_r($result);
 echo json_encode($result);

 $final=array();
 $sql_courses=mysql_query("SELECT course_id, course_name FROM course") or die (mysql_error());
 $row_courses = mysql_fetch_array($sql_courses);            
 $result=$row_courses['course_name']; // this will have first_coures_name (an string)
 $final['cname']=$result;
 //print_r($result);
 echo json_encode($final); //the output should be this {'cname':'first_course_name'}

为什么在您不使用用户id时发送用户id？它是否有任何值？什么是
alert（msg）显示？您也没有在PHP中的任何地方使用
user\u id
。@user1657720您应该接受您喜欢的答案。您应该单击要接受的答案左侧的复选标记。