[PHP]将JSON模型的数据，替换成JSON数据

Question

{data:[{code:"现货黄金",newPrice:"1662.3",hight:"0",low:"0",buy:"1662.3",sell:"r1663.3",zhang:"r4.75",date:"4:59:40",open:"0",close:"1657.55"}, {code:"香港黄金",newPrice:"15384",hight:"0",low:"0",buy:"15384",sell:"r15390",zhang:"r44",date:"2:59:55",open:"0",close:"15340"}, {code:"现货白银",newPrice:"31.25",hight:"0",low:"0",buy:"31.25",sell:"r31.31",zhang:"r0.17",date:"4:56:10",open:"0",close:"31.08"}, {code:"现货白金",newPrice:"1566.75",hight:"0",low:"0",buy:"1566.75",sell:"r1576.75",zhang:"r2.01",date:"3:29:35",open:"0",close:"1564.74"}, {code:"现货钯金",newPrice:"677.23",hight:"0",low:"0",buy:"677.23",sell:"r684.23",zhang:"r9.54",date:"3:29:35",open:"0",close:"667.69"}]}
大家看好了，这是具有一个JSON数据的模型，只是键名没有引号，所以json_decode会认为它是非法的JSON数据格式。现要用正则将它替换成，且加引号。/(?<={|,)[^:\s]*/is，目前能match出一个二维数据，但replace的话，不行呀，必须得是一个一维数组。请教各位大侠。

Answer 1

<?php
$json=<<<JSON
{data:[{code:"现货黄金",newPrice:"1662.3",hight:"0",low:"0",buy:"1662.3",sell:"r1663.3",zhang:"r4.75",date:"4:59:40",open:"0",close:"1657.55"}, {code:"香港黄金",newPrice:"15384",hight:"0",low:"0",buy:"15384",sell:"r15390",zhang:"r44",date:"2:59:55",open:"0",close:"15340"}, {code:"现货白银",newPrice:"31.25",hight:"0",low:"0",buy:"31.25",sell:"r31.31",zhang:"r0.17",date:"4:56:10",open:"0",close:"31.08"}, {code:"现货白金",newPrice:"1566.75",hight:"0",low:"0",buy:"1566.75",sell:"r1576.75",zhang:"r2.01",date:"3:29:35",open:"0",close:"1564.74"}, {code:"现货钯金",newPrice:"677.23",hight:"0",low:"0",buy:"677.23",sell:"r684.23",zhang:"r9.54",date:"3:29:35",open:"0",close:"667.69"}]}
JSON;$json=preg_replace_callback('/(,|,?\s*?{)\b(\w+):/','callback',$json);
print_r(json_decode($json,true));function callback($v){
return "{$v[1]}\"{$v[2]}\":";
}正常输出数组 应该是你要的结果

Answer 2

楼上的，非常棒，是我要的结果。想请问一下：用正则(?<={|,)[^:\s]*(?=:)这样来，不行吗？？

Answer 3

$json=preg_replace('/(,|,?\s*?{)\b(\w+):/', '$1"$2":',$json);
print_r(json_decode($json,true));

Answer 4

当然可以$json=preg_replace('/(?<={|,)[^:\s]*(?=:)/is', '"$0"',$json);
print_r(json_decode($json,true));

Answer 5

你们写的都比我写的效率高，我的要回塑，效率大大折扣。$pattern = '/(?<={|,)[^:\s]*(?=:)/is';
$json=preg_replace_callback($pattern, 'callk',$str);function callk($v){
return "\"$v[0]\"";
}
echo $json;
print_r(json_decode($json,true));exit;我是参考了@Mr_merlin的写法，将我自个的改了下，也能用。不过，@xuzuning你的这个效率应该更高。