Php 从xmlhttp.responseText使用JSON解析数组

我正在尝试使用ajax将php数组转换为js，并在php中使用JSON编码，在js中解码，但无法获得所需的结果

我在回显编码字符串之前使用var_dump，这是输出：

$boardArrayEncoded=json\u encode（$boardArray）；
var_dump（$boardArrayEncoded）；
echo$boardArrayEncoded；

JSON_encode后面的字符串是：

string '{"1x1":0,"1x2":0,"1x3":0,"1x4":0,"1x5":0,"1x6":0,"2x1":0,"2x2":0,"2x3":0,"2x4":0,"2x5":0,"2x6":0,"3x1":0,"3x2":0,"3x3":0,"3x4":0,"3x5":0,"3x6":0,"4x1":0,"4x2":0,"4x3":0,"4x4":0,"4x5":0,"4x6":0,"5x1":0,"5x2":0,"5x3":0,"5x4":0,"5x5":0,"5x6":0,"6x1":0,"6x2":0,"6x3":0,"6x4":0,"6x5":0,"6x6":0}'

在javascript中，我尝试解析它：

if（xmlhttp.readyState==4&&xmlhttp.status==200）
{
document.getElementById（“logDiv1”）.innerHTML=xmlhttp.responseText；
array=JSON.parse（xmlhttp.responseText）；
}

这是一个xmlhttp。回应文本：：<代码>{“1x1:1:0”，1x3:0，“1x4:0”，1x5:0，“1x5:0”，1x6:0，“1x6”：0”，1x1:0，“1x1:0”，1x1:0，“1x1:0”，1x1:0，“1x1:0”，1x1:0，“1x3:0”，1x3:0，“1x5:0”，1x5:0，“1x5:0”，1x5:0，“1x5:0”，0，“1x5:0”，0，“0”，0，“4x0，”4x0，”4x3:0，“4x0，”4x10，”4x10，“4x10，”4x0，”0，”4x0，“4x0，”4x0，“4x0”，0，”4x0，“4x0”，0，”4x0，“4x0”，0，”0，”4x0，“4x0”，0，“4x0，“4x3”0，”“6x5”：0，“6x6”：0}

在firebug上，我发现语法错误：

SyntaxError: JSON.parse: unexpected character
[Break On This Error]   

array1 = JSON.parse(xmlhttp.responseText);

我做错了什么？我需要在JS中使用这个数组，如何正确编码它

---

提前感谢

删除var\u转储。它是打印输出，这可能导致您所说的responseText不是完整的responseText

我希望你的答复如下：

string '{"1x1":0,"1x2":0,"1x3":0,"1x4":0,"1x5":0,"1x6":0,"2x1":0,"2x2":0,"2x3":0,"2x4":0,"2x5":0,"2x6":0,"3x1":0,"3x2":0,"3x3":0,"3x4":0,"3x5":0,"3x6":0,"4x1":0,"4x2":0,"4x3":0,"4x4":0,"4x5":0,"4x6":0,"5x1":0,"5x2":0,"5x3":0,"5x4":0,"5x5":0,"5x6":0,"6x1":0,"6x2":0,"6x3":0,"6x4":0,"6x5":0,"6x6":0}'
{"1x1":0,"1x2":0,"1x3":0,"1x4":0,"1x5":0,"1x6":0,"2x1":0,"2x2":0,"2x3":0,"2x4":0,"2x5":0,"2x6":0,"3x1":0,"3x2":0,"3x3":0,"3x4":0,"3x5":0,"3x6":0,"4x1":0,"4x2":0,"4x3":0,"4x4":0,"4x5":0,"4x6":0,"5x1":0,"5x2":0,"5x3":0,"5x4":0,"5x5":0,"5x6":0,"6x1":0,"6x2":0,"6x3":0,"6x4":0,"6x5":0,"6x6":0}

检查内容类型：

  header('Content-type: application/json');

---

确保您只有

echo$boardArrayEncoded和无其他输出。还要确保没有前导空格或换行符，因为这也会打断有效的JSON数据。在执行echo之前，包括
header（'Content-type:application/json'）以JSON格式准备输出。

“1x5”：0，“3x6”0，“3x6”0，“3x6”0，“3x6”0，“3x6”0，“4x1”：0，“1x6”：0，”1x6”：0，“1x1”：0，“2x1”：0，“1x1”：0，“1x1”：0，”1x1：0，“1x6”：0，“1x5”0，“1x5”：0，“1x5”0”，1x6：0，“4x1”：0，”0，“4x3”：0，“4x3”0，”4x3”0，“4x3”：0，”0，“4x3”0，“4x3”0，”4x3”0，“4x3”：0，“4x3”0，”4x3”0，“4x3”0，”4x3”0，“4x3”0，”4x3”0，“4x3”0，“4x3”0，“4x3”：0，“4x3”0，“4x3”0，”4x3”0，”：0，“6x6”：0}'；//您的JSON字符串。
parsedJSON=eval（'（'+responseText+'））；//解析的JSON。对象
警报（parsedJSON[“1x1”]）；//访问对象元素是这样的，因为您不能编写parsedJSON.1x1；
for（parsedJSON中的元素）{
document.body.innerHTML+=element+“=>”+parsedJSON[element]+“
”；
}//只是一个vardump，表示一切都很完美
根据，您发布的JSON是有效的。你确定这就是你得到的吗？是的，这是从var_dump和innerHTML打印出来的所有输出。好吧，如果
string
是响应的一部分（如果你的代码中真的有
var_dump
），那么这就是问题所在。我尝试了JSON.parse（上面的响应文本），它没有给我任何错误，它完成了任务，我删除了var_转储及其工作，可能问题是var_转储正在写入responseText，然后响应就像您在回答中提到的那样，阻止了它使用正确的JSON格式。
var responseText='{"1x1":0,"1x2":0,"1x3":0,"1x4":0,"1x5":0,"1x6":0,"2x1":0,"2x2":0,"2x3":0,"2x4":0,"2x5":0,"2x6":0,"3x1":0,"3x2":0,"3x3":0,"3x4":0,"3x5":0,"3x6":0,"4x1":0,"4x2":0,"4x3":0,"4x4":0,"4x5":0,"4x6":0,"5x1":0,"5x2":0,"5x3":0,"5x4":0,"5x5":0,"5x6":0,"6x1":0,"6x2":0,"6x3":0,"6x4":0,"6x5":0,"6x6":0}'; // Your JSON string.
parsedJSON=eval('('+responseText+')'); // Parsed JSON. Object.
alert(parsedJSON["1x1"]); // Access object element's like this, because you can't write parsedJSON.1x1;
for(element in parsedJSON){
    document.body.innerHTML+=element+" => "+parsedJSON[element]+"<br />";
} // Just a vardump to show everything's perfect