Question

我试图通过使用php将mysql数据引入jquery，我将数据转换为这样的JSON格式。

{"uid":"33","title":"Apple, Peach, Grapefruit","ing1":"apple","qty1":"1","meas1":"whole","ing2":"peaches \/ halved and","qty2":"2","meas2":"each","ing3":"grapefruit \/ peeled","qty3":"2","meas3":"each","ing4":"","qty4":"0","meas4":"each","ing5":"","qty5":"0","meas5":"each","ing6":"","qty6":"0","meas6":"each","ing7":"","qty7":"0","meas7":"each","ing8":"","qty8":"0","meas8":"each","ing9":"","qty9":"0","meas9":"each","ing10":"","qty10":"0","meas10":"each","servings":"2","benefits":""}

使用以下代码：

require_once'connect.php'; 
$uid = $_GET['uid'];
$sql = "SELECT * FROM recipes WHERE uid = '$uid'";
$result = mysqli_query($conn, $sql);

$num_rows = mysqli_affected_rows($conn);
while($row = mysqli_fetch_assoc($result)) {
$data = json_encode($row);
echo $data;
}

我正在使用jquery .get将此代码拉入网页。

$(document).ready(function(e) {
var id = location.search;
uid=id.substring(4);
$.get('../jqm_juicing/data/get_json.php?uid=' + uid,function(data, status){
$("#display").append(data);
});
});

它显示如上所述的json数据。我希望能够单独访问不同的元素，我该怎么做？

Answer 1

在jQuery中，在追加（数据）之前添加JSON.parse（）。

$(document).ready(function(e) {
    var id = location.search;
    uid=id.substring(4);
    $.get('../jqm_juicing/data/get_json.php?uid=' + uid,function(data, status){
        $("#display").append(JSON.parse(data));
    });
});

我刚刚更改了这一行：

$("#display").append(JSON.parse(data));

希望这有帮助！

Answer 2

如果用<script>标签包装它，并将其声明为变量，则可以使用它。

require_once'connect.php'; 
$uid = $_GET['uid'];
$sql = "SELECT * FROM recipes WHERE uid = '$uid'";
$result = mysqli_query($conn, $sql);

$num_rows = mysqli_affected_rows($conn);

echo "<script>";
while($row = mysqli_fetch_assoc($result)) {
  $data = json_encode($row);
  echo "var myJson=" . $data;
}
echo "</script>";

现在在您的主页脚本中，您可以像任何其他普通JSON一样访问此JSON 以下是您可以在主页脚本中的任何位置放置的间隔：
只是为了测试它;）

var checkJson = setInterval(function(){

  if(typeof(myJson.uid)!="undefined"){

    clearInterval(checkJson);

    console.log("myJson uid: " + myJson.uid);
    console.log("myJson title: " + myJson.title);
    // etc...

  }else{
    console.log("JSON not loaded yet.");
  }

},500);

Answer 3

检索json数据以使用json response jquery parse函数。使用此函数的json数据 jQuery.parseJSON（）

http://api.jquery.com/jquery.parsejson/

从mysql和php检索json数据到jquery

3 个答案: