2015-08-14 71 views
1

我想做一個Ajax調用,並從PHP腳本接收json,然後將在js代碼中處理。JQuery沒有接收json數據結構只是empyt []

因爲當JSON傳遞給回調函數的空數組某種原因...所以console.log(json)[]

然而,當我打電話PHP腳本它給了我有效的JSON作爲的["7494","7495","7565"]輸出瀏覽器。

這是AJAX/JQuery的:

$.ajax({ 
     url:'validation_ajax.php', 
     data:{ 
     'jobids': 'yes', 
     'projectid': '3062' 
     }, 
     type: 'GET', 
     dataType: 'json', 
    success: function(d){ 
       console.log(d); 
    }, 
    error: function(xhr, status, errorThrown) { 
        alert("Sorry, there was a problem!"); 
        console.log("Error: " + errorThrown); 
        console.log("Status: " + status); 
        console.dir(xhr); 
    } 
    }); 

的PHP是這樣

$projectId = testInput($_GET['project']); 
$con = dbConnect(); 
$sql = "select distinct iJobID from Validation_Samples 
     where iProjectID = $projectId"; 
$result = mysqli_query($con, $sql); 

$jobs = array(); 
    while($row = mysqli_fetch_array($result)){ 

     $jobs[] = $row['iJobID']; 
    } 
    echo json_encode($jobs); 

    mysqli_close($con); 
+0

相關:http://stackoverflow.com/questions/5147522/jquery-ajax-request-of-datatype-json-will-not-retrieve-data-from-php-script – Und3rTow

回答

1

$_GET['project']是空的,應該是$_GET['projectid']我承擔。 回聲之前還要做mysqli_close($con)

+0

感謝您的密切關注。 – SciGuyMcQ