获取PHP中生成的每一行JSON数据

这是一个艰难的过程，至少对我来说是这样。所以基本上我想做的是通过Javascript获取JSON数据每行的每一个值：

{"id":2,"url":"image.png","x":19,"y":10,"user_id":20}
{"id":3,"url":"image.png","x":19,"y":10,"user_id":20}
{"id":4,"url":"image.png","x":19,"y":10,"user_id":20}
{"id":5,"url":"image.png","x":19,"y":10,"user_id":20}
{"id":6,"url":"image.png","x":19,"y":10,"user_id":20}
{"id":7,"url":"image.png","x":19,"y":10,"user_id":20}
{"id":8,"url":"image.png","x":19,"y":10,"user_id":20}
{"id":9,"url":"image.png","x":19,"y":10,"user_id":20}
{"id":10,"url":"image.png","x":19,"y":10,"user_id":20}
{"id":31,"url":"image.png","x":19,"y":10,"user_id":20}
{"id":32,"url":"image.png","x":19,"y":10,"user_id":20}
{"id":33,"url":"image.png","x":19,"y":10,"user_id":20}
{"id":34,"url":"image.png","x":19,"y":10,"user_id":20}
{"id":35,"url":"image.png","x":19,"y":10,"user_id":20}
{"id":36,"url":"image.png","x":19,"y":10,"user_id":20}
{"id":37,"url":"image.png","x":19,"y":10,"user_id":20}
{"id":38,"url":"image.png","x":19,"y":10,"user_id":20}
{"id":39,"url":"image.png","x":19,"y":10,"user_id":20}
{"id":40,"url":"image.png","x":19,"y":10,"user_id":20}

我是这样生成的（在我的PHP文件中）：

我已经尝试了很多与jQuery的

$相关的代码。每一个都是
，但是运气不好，代码永远不会工作。我希望得到每个值，因为我将在页面上显示它们，位置和图像都正确。有人知道获取每一行的正确函数吗？就像PHP的
$row['foo']

-----编辑------

我是这样做的：

$.get('/Application/Ajax/__ajaxProfile.php?a=GetWidgets', function(data) {

var json = data;

for (i=0;i<json.length;i++){
    var obj = json[i];
    for (var key in obj) {
        document.getElementById("log").innerHTML += key+": "+obj[key]+"<br />";
    }
    document.getElementById("log").innerHTML += "<br />";
} 

});

for (i=0;i<json.length;i++){
  var obj = json[i];
  for (var key in obj) {
    document.write(key+": "+obj[key]);
  }
}

$.get（'/Application/Ajax/\uuu ajaxProfile.php？a=GetWidgets'，函数（数据）{
var json=数据；
对于（i=0；i这可能是您想要的

$db->where('user_id', '20');
$results = $db->get('profile_stickers');

$arr = array();

foreach ($results as $parameters => $values) {

    $arr[] = $values;

}

echo json_encode($arr);

这可能是你想要的

$db->where('user_id', '20');
$results = $db->get('profile_stickers');

$arr = array();

foreach ($results as $parameters => $values) {

    $arr[] = $values;

}

echo json_encode($arr);

这可能不是您想要的，但这是一种在javascript中完成一切的方法

您基本上需要将其存储为JSON对象数组：

var json = [
{"id":2,"url":"image.png","x":19,"y":10,"user_id":20},
{"id":3,"url":"image.png","x":19,"y":10,"user_id":20},
...etc
]

然后，您可以像这样在阵列和对象之间循环：

$.get('/Application/Ajax/__ajaxProfile.php?a=GetWidgets', function(data) {

var json = data;

for (i=0;i<json.length;i++){
    var obj = json[i];
    for (var key in obj) {
        document.getElementById("log").innerHTML += key+": "+obj[key]+"<br />";
    }
    document.getElementById("log").innerHTML += "<br />";
} 

});

for (i=0;i<json.length;i++){
  var obj = json[i];
  for (var key in obj) {
    document.write(key+": "+obj[key]);
  }
}

for（i=0；i这可能不是您想要的，但这是一种在javascript中完成一切的方法

您基本上需要将其存储为JSON对象数组：

var json = [
{"id":2,"url":"image.png","x":19,"y":10,"user_id":20},
{"id":3,"url":"image.png","x":19,"y":10,"user_id":20},
...etc
]

然后，您可以像这样在阵列和对象之间循环：

$.get('/Application/Ajax/__ajaxProfile.php?a=GetWidgets', function(data) {

var json = data;

for (i=0;i<json.length;i++){
    var obj = json[i];
    for (var key in obj) {
        document.getElementById("log").innerHTML += key+": "+obj[key]+"<br />";
    }
    document.getElementById("log").innerHTML += "<br />";
} 

});

for (i=0;i<json.length;i++){
  var obj = json[i];
  for (var key in obj) {
    document.write(key+": "+obj[key]);
  }
}

用于（i=0；这不是一个有效的JSON。从PHP生成JSON时，您应该只调用
JSON\u encode
一次！生成您想要的结构，然后在最后调用
JSON\u encode
。这不是一个有效的JSON。从PHP生成JSON时，您应该只调用
JSON\u encode
一次！生成您想要的结构，然后在最后调用
json\u encode
。