Question

我将JSON对象动态创建为：

[{"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追加到单个Javascript对象中：

[{"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，并将它放在一个不同的Javascript变量中：

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，我希望在同一个变量中生成所有新生成的JSON obj。谁能告诉我怎么办呢？

Answer 1

在将对象添加到主阵列之前，您需要将对象拉出各个阵列：

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

显然，对于n个对象，你将循环使用它，而不是像上面那样。

在Javascript中附加JSON对象

1 个答案: