2014-05-11 31 views
1

我的問題是,我不能檢索通過ajax MySQL的結果的結果,請大家幫忙檢索的MySQL結果將jQuery

Ajax代碼:

$.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); 
    } 
}); 

PHP代碼:

$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); 
+1

爲什麼你在不使用它的時候發送user_id,它有什麼價值嗎? –

+0

'alert(msg);'show?你也不會在PHP的任何地方使用'user_id'。 –

回答

1

更改PHP代碼如下

$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); 

變化PHP代碼如下:

而且它能夠更好地使用console.log(variable);來檢查變量的內容

Ajax代碼:

$.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); 
    } 
}); 
0

其更好地與這樣的鍵值發送:

$.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代碼:

$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'} 
+0

你好,tkx所有,它的工作......非常感謝你 – user1657720

+0

@ user1657720你應該接受你喜歡的答案。您應該點擊您想要接受的答案左側的複選標記。 – ncm