php - 无法通过从 php 文件传递​​ JSON 数据来使用 jquery 数据表填充数据表

Question

我正在尝试使用来自 Jquery 的数据表插件并从 PHP 传递 JSON 数据来填充数据表。但是我收到无效的 JSON 错误。我已经使用 JSON lint 验证了我的 json，它成功验证了它

我的代码如下所示：

//SQL query to get data from Database

$sql_fabentry="select f.fabrication_date as date,concat(w.workstage_name,':',f.fabricator_remarks)"
            . " as workstage_entry,u.user_name from fabrication_record f,workstage w,"
            . "users u where f.workstage_id=w.workstage_id and u.user_id=f.fabricator_id "
            . "and f.subsystem_id='$subsystem_id_session' and f.cp_no IS NULL";


    $sql_qcentry="select f.qc_date as date,concat(r.remark,':',f.qc_remarks)"
            . " as workstage_entry,u.user_name from fabrication_record f,remarks r,"
            . "users u where f.qcremarks_id=r.remark_id and u.user_id=f.qcinspector_id "
            . "and f.subsystem_id='$subsystem_id_session' and f.cp_no IS NULL";

//db_select function return an associative array
//result from both queries will always return same number of rows
$results_fabentry=$db->db_select($sql_fabentry);
$results_qcentry=$db->db_select($sql_qcentry);

 //Here iam trying to combine the two arrays in one array with alternating 
 //rows from each array

$count1=count($results_fabentry);
$i=0;
$j=0;
$result=array();
while($i<$count1){
    $result[$j]['date']=$results_fabentry[$i]['date'];
    $result[$j]['workstage_entry']=$results_fabentry[$i]['workstage_entry'];
    $result[$j]['user_name']=$results_fabentry[$i]['user_name'];

    $result[$j+1]['date']=$results_qcentry[$i]['date'];
    $result[$j+1]['workstage_entry']=$results_qcentry[$i]['workstage_entry'];        
    $result[$j+1]['user_name']=$results_qcentry[$i]['user_name'];
    $i++;
    $j+=2;
}


$results=["sEcho=>1",
            "iTotalRecords"=>count($result),
            "iTotalDisplayRecords"=>($result),
            "aaData"=> $result];

echo json_encode($results);

我已经验证了 $ result 中的 JSON 数据并且它是有效的。如果删除 while 循环并尝试遵循它的工作原理：

$results=["sEcho=>1",
            "iTotalRecords"=>count($results_fabentry),
            "iTotalDisplayRecords"=>($results_fabentry),
            "aaData"=> $results_fabentry];

echo json_encode($results);

谁能解释我做错了什么？

标签： phpjquery

Answer 1

我能够通过用 foreach 循环替换 while 循环来解决问题。我使用了以下代码并且工作正常

$result = [];

$i=0;
$j=1;
foreach($results_fabentry as $result_fabentry){
    $result[$i]['date']=$result_fabentry['date'];
    $result[$i]['workstage_entry']=$result_fabentry['workstage_entry'];
    $result[$i]['user_name']=$result_fabentry['user_name'];
    $result[$i+1]['date']="0";
    $result[$i+1]['workstage_entry']="0";
    $result[$i+1]['user_name']="0";

    $i+=2;
}

foreach($results_qcentry as $result_qcentry){

    $result[$j]['date']=$result_qcentry['date'];
    $result[$j]['workstage_entry']=$result_qcentry['workstage_entry'];        
    $result[$j]['user_name']=$result_qcentry['user_name'];

    $j+=2;
}

$results=["sEcho=>1",
            "iTotalRecords"=>count($result),
            "iTotalDisplayRecords"=>($result),
            "aaData"=> $result];

echo json_encode($results);