在Javascript中附加JSON对象

我将动态创建JSON对象，如下所示：

[{"fill":"none","stroke":"#000000","path":"M186.5,25L187.5,25L187.5,26L188.5,27L189.5,28L189.5,29L190.5,29","stroke-opacity":1,"stroke-width":5,"stroke-linecap":"round","stroke-linejoin":"round","transform":[],"type":"path"}]

[{"fill":"none","stroke":"#000000","path":"M73.5,42L73.5,42L75.5,43L82.5,46L101.5,55L119.5,65L126.5,69L128.5,71L129.5,71","stroke-opacity":1,"stroke-width":5,"stroke-linecap":"round","stroke-linejoin":"round","transform":[],"type":"path"}]

.......

我想将生成的所有这些对象附加到单个Javascript对象中，如下所示：

[{"fill":"none","stroke":"#000000","path":"M186.5,25L187.5,25L187.5,26L188.5,27L189.5,28L189.5,29L190.5,29","stroke-opacity":1,"stroke-width":5,"stroke-linecap":"round","stroke-linejoin":"round","transform":[],"type":"path"},
{"fill":"none","stroke":"#000000","path":"M73.5,42L73.5,42L75.5,43L82.5,46L101.5,55L119.5,65L126.5,69L128.5,71L129.5,71","stroke-opacity":1,"stroke-width":5,"stroke-linecap":"round","stroke-linejoin":"round","transform":[],"type":"path"}]

这样，创建的每个对象都应该附加到此JSON字符串。 我能够连接两个JSON对象，并将其放在不同的Javascript变量中，如下所示：

var obj1 = '[{"fill":"none","stroke":"#000000","path":"M186.5,25L187.5,25L187.5,26L188.5,27L189.5,28L189.5,29L190.5,29","stroke-opacity":1,"stroke-width":5,"stroke-linecap":"round","stroke-linejoin":"round","transform":[],"type":"path"}]';

var obj2 = '[{"fill":"none","stroke":"#000000","path":"M186.5,25L187.5,25L187.5,26L188.5,27L189.5,28L189.5,29L190.5,29","stroke-opacity":1,"stroke-width":5,"stroke-linecap":"round","stroke-linejoin":"round","transform":[],"type":"path"}]';

var mergedJS = JSON.parse(obj1).concat(JSON.parse(obj2));

  mergedJSON =JSON.stringify(mergedJS);

但是，我希望所有新生成的JSON objs都在同一个变量中。

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有人能告诉我怎么做吗？

在将对象添加到主阵列之前，您需要将对象从各自的阵列中拉出：

var newJSArray = [];
var mergedJS = JSON.parse(obj1);
newJSArray.push(mergedJS[0]);
mergedJS = JSON.parse(obj2);
newJSArray.push(mergedJS[0]);

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显然，对于n个对象，您将循环该对象，而不是像我上面所做的那样。

这些对象是如何生成的？不能直接将它们推送到数组中吗？我不能直接将它们推送到数组中，因为它们是在函数调用中单独生成的。如果可能的话，我如何将它们直接推送到数组中呢？但是这个函数会被调用，它会返回对象吗？这些对象生活在哪里？使用

var mergedJS=JSON.parse（obj1）.concat（JSON.parse（obj2））是，这些对象被创建为$（“#data2”）.val（sketchpad.json（））；和被分配到一个隐藏的div，ID=“data2”我希望此时将创建的每个对象添加到数组中。这段代码将完成相同的任务。只需获取每个对象数组的第0个元素，因为您的对象是在封闭数组中发送的。