jquery - 无法通过ajax调用在framework7中获取php解析的数据

Question

function pendingorder(){

            app.request({
            type:"POST",
            url: "pages/getpeningorer.php",
            dataType: 'json',
            cache: false,
            success:function(data) {
                console.log(data);
             var result = $.parseJSON(data);
              $.each(result, function(key, value){
                $.each(value, function(k, v){
                    if(k === "order_id"){
                        $("#pendingtable >tbody:last").append(
                            $('<tr>').append(
                                $('<td>').append(v)
                                .append(
                                    $('</td>').append(
                                        $('</tr>')
                                        )
                                    )
                                )
                            );
                    }
                    if(k === "product_id"){
                        $("#demoTable >tbody >tr:last").append(

                            $('<td>').append(v)
                            .append(
                                $('</td>')

                                )

                            );
                    }

                    if(k === "status"){
                        $("#demoTable >tbody >tr:last").append(

                            $('<td>').append(v)
                            .append(
                                $('</td>')

                                )

                            );
                    }

                    if(k === "remark"){
                        $("#demoTable >tbody >tr:last").append(

                            $('<td>').append(v)
                            .append(
                                $('</td>')

                                )

                            );
                    }
                    if(k === "postingDate"){
                        $("#demoTable >tbody >tr:last").append(

                            $('<td>').append(v)
                            .append(
                                $('</td>')

                                )

                            );
                    }

                    });
            });
        console.log(data);}
    });
      console.log('execute success');   
}

 I AM trying to call ajax through function....But not working. In a similar way, I post data it is working.

PHP代码：

在此处输入代码

$conn = new mysqli($host, $dbUsername, $dbPassword, $dbname);

    if (mysqli_connect_error()) {
     die('Connect Error('. mysqli_connect_errno().')'. mysqli_connect_error());
    } else {
     $sql = "SELECT order_id,product_id,status,remark,postingDate FROM order_track_history where status='In process'";
     $result = $conn->query($sql);
    }
    $Pdata = array();
    while ($row = mysql_fetch_array($result)) {
    $picture = array(
    "order_id" => $row['order_id'],
    "product_id"         => $row['product_id'],
    "status"          => $row['status'],
    "remark"       => $row['remark'],
    "postingDate"       => $row['postingDate']
  );
  $Pdata[] = $picture;
}`enter code here`
    echo json_encode($Pdata);

在这里，我将我的数据以 JSON 格式发送到 ajax 调用。但无法在 HTML 页面上看到数据。

标签： jquerymysqlajaxhtml-framework-7

Answer 1

您的 php 代码似乎有问题，请尝试使用此代码

$row = mysql_fetch_array($result);
foreach($row as $r) {
    $picture = array(
    "order_id" => $r['order_id'],
    "product_id"         => $r['product_id'],
    "status"          => $r['status'],
    "remark"       => $r['remark'],
    "postingDate"       => $r['postingDate']
    );
}