Question

我目前正在使用Laravel 5。在我的GoalsController中：

public function getGoalData()
{
    // my queries
    return view('userPages.usercp')->with('goal_data', $goal_data);
}

当我按$goal_data打印<?php echo $goal_data ?>时，它是一个json字符串。现在我想通过使用Javascript进一步处理来阅读它。我怎样才能做到这一点？谢谢。

编辑

我想在外部脚本中读取@goal_data。以下是Julian Paolo Dayag和VMcreator的解决方案。

在您的HTML文件中，将其放在最后（</body>标记之前）：

<script> window.GOALS = <?echo json_encode($goal_data); ?> </script>

根据Julian Paolo Dayag的说法：

通过将变量存储到窗口对象中，意味着它将变为a 全局变量。

现在在您的JS文件中，您可以这样调用它：var goalData = JSON.parse(GOALS);

Answer 1

在javascript中使用JSON.parse将JSON字符串解析为Object。

JSON.parse(text[, reviver])

请在此处阅读：https://developer.mozilla.org/en/docs/Web/JavaScript/Reference/Global_Objects/JSON/parse

编辑

你可以这样做：

<script>
     var goal_data=JSON.parse(<?php echo $goal_data ?>);
</script>

如果它在laravel的刀片中，你可以使用JSON.parse({{ $goal_data }});

修改

我尝试了这样的示例JSON字符串：

<?php $goal='[ { "goalId":28, "title":"1st goal", "description":"what the fuck", "startTime":"00:00:00", "endTime":"22:00:00", "remind":0, "autocheck":1, "repeat":"once", "goalStatus":"success", "type":"daily", "freqYear":0, "start":"08\/13\/2015", "end":"" }, { "goalId":29, "title":"2nd goal", "description":"Lorem ipsum dolor sit amet", "startTime":"17:25:00", "endTime":"17:25:00", "remind":1, "autocheck":0, "repeat":"once", "goalStatus":"success", "type":"daily", "freqYear":0, "start":"08\/13\/2015", "end":"" }, { "goalId":30, "title":"3rd goal", "description":"Lorem ipsum dolor sit amet, consectetur adipiscing elit", "startTime":"05:00:00", "endTime":"14:00:00", "remind":1, "autocheck":1, "repeat":"until", "goalStatus":"success", "type":"daily", "freqYear":0, "start":"08\/13\/2015", "end":"08\/19\/2015" }, { "goalId":31, "title":"4th goal", "description":"Lorem ipsum dolor sit amet, consectetur adipiscing elit", "startTime":"00:00:00", "endTime":"16:00:00", "remind":0, "autocheck":1, "repeat":"once", "goalStatus":"success", "type":"weekly", "freqYear":2015, "start":"1", "end":"" }, { "goalId":32, "title":"5th goal", "description":"Lorem ipsum dolor sit amet, consectetur adipiscing elit", "startTime":"00:00:00", "endTime":"00:00:00", "remind":1, "autocheck":1, "repeat":"until", "goalStatus":"success", "type":"weekly", "freqYear":2015, "start":"9", "end":"14" }, { "goalId":33, "title":"6th goal", "description":"Lorem ipsum dolor sit amet, consectetur adipiscing elit", "startTime":"00:00:00", "endTime":"00:00:00", "remind":1, "autocheck":1, "repeat":"once", "goalStatus":"success", "type":"monthly", "freqYear":2018, "start":"8", "end":"" }, { "goalId":34, "title":"7th goal", "description":"Lorem ipsum dolor sit amet, consectetur adipiscing elit", "startTime":"00:00:00", "endTime":"20:00:00", "remind":0, "autocheck":1, "repeat":"until", "goalStatus":"success", "type":"yearly", "freqYear":0, "start":"2015", "end":"2029" } ]'; ?> <script type="text/javascript"> console.log(JSON.parse(<?php echo json_encode($goal) ?>)); </script>

没有错误。

修改

如果你想将你的php变量传递给外部JS，你可以将它分配给全局变量，在你的head标签中，你的代码应该是这样的：

..assuming this is your head tag <script> var goal_string=<?php echo json_encode($goal_data); ?>; </script> <script src="external_js.js" ></script>

现在您可以访问external_js.js上的var goal_string以及其声明下面包含的任何JS。

Answer 2

您可以使用JSON.parse("json string here");

示例：

单个对象

    var person = JSON.parse("{name: 'hello world', age: 18}");

    console.log(person.name); // outputs "hello world"
    console.log(person.age);  // outputs 18

对象数组

    var person = JSON.parse("[{name: 'hello world', age: 18},{name: 'foo bar', age: 10}]");

    console.log(person[0].name); // outputs "hello world"
    console.log(person[0].age);  // outputs 18
    console.log(person[1].name); // outputs "foo bar"
    console.log(person[1].age);  // outputs 10

有时 JSON字符串出错，导致 JSON.parse（）在解析失败时抛出异常。

建议将 JSON.parse（）包装在 try / catch块周围，以便javascript不会停止执行您的脚本。

    try {
        var person = JSON.parse("[{name: 'hello world', age: 18},{name: 'foo bar', age: 10}]");

        console.log(person[0].name); // outputs "hello world"
        console.log(person[0].age);  // outputs 18
        console.log(person[1].name); // outputs "foo bar"
        console.log(person[1].age);  // outputs 10
    } catch(e) {
        console.log("JSON parse failure!");
    }

Answer 3

有各种方法可以使用JS： -

var person = JSON.parse("[{JSON1: 'Value1', JSON2: value2},{JSON1: 'Value1', JSON2: Value2}]");
console.log(person[0].JSON1);
console.log(person[0].JSON2);
alert(person[0].JSON1);
alert(person[0].JSON2);

使用Java代码：

 String s = "[0,{\"1\":{\"2\":{\"3\":{\"4\":[5,{\"6\":7}]}}}}]";
 Object obj = parser.parse(s);
 JSONArray array = (JSONArray)obj;
 System.out.println(array.get(1));

或在脚本标记内的PHP访问文件中： -

 var goal_data=JSON.parse("<?php echo $goal_data ?>");

如何从JSON响应中读取数据

3 个答案: