Question

我通过api在数据库中保存了一些值（所以我不能手动修改它）

当我从数据库返回时，json值不是有效的形式。我不想修改数据库中的每个值。我想在php中使用可以将其转换为有效格式的内容。

我的数据库中有这样的东西：

$invalid='{
    "response": {

                "id": "16"",  <--------------------- Invalid Format(It can be for any key)
                "event_name": "testing",
                "image": "images/Penguins.jpg",
                "event_date": "2014-12-13",
                "event_time": "02:10",
                "time_interval": "4",
                "location": "mohali",
                "event_type": "Rock",
                "detail": "sfdsf fgf ghb\t",
                "delivery": "dggh fghgfh\t\t",
                "status": "1"
            }
           }';

那么如何通过php将其转换为有效的json：

$invalid='{
    "response": {

                "id": "16",  <--------------------- valid Format
                "event_name": "testing",
                "image": "images/Penguins.jpg",
                "event_date": "2014-12-13",
                "event_time": "02:10",
                "time_interval": "4",
                "location": "mohali",
                "event_type": "Rock",
                "detail": "sfdsf fgf ghb\t",
                "delivery": "dggh fghgfh\t\t",
                "status": "1"
            }
           }';

Answer 1

对于这个确切的情况，这将做：

<?php
$invalid='{
  "response": {        
    "id": "16"", 
    "event_name": "testing",
    "image": "images/Penguins.jpg",
    "event_date": "2014-12-13",
    "event_time": "02:10",
    "time_interval": "4",
    "location": "",
    "event_type": "Rock"",
    "detail": "sfdsf fgf ghb\t",
    "delivery": "dggh fghgfh\t\t",
    "status": "1""
  },
  "bar":{ "test": ""
  }
}';
$invalid = explode("\n",$invalid);
foreach($invalid as $idx => &$line) {
  $num = 0;
  for($z=0;$z<strlen($line);++$z) {
    $ch = $line[$z];
    if($ch == "\\") {
      ++$z;
    } else if($ch == '"') {
      ++$num;
      $last_pos = $z;
    }
  }
  if($num % 2 == 1) {
    $line[$last_pos] = ' ';
  }             
}       
$valid = implode("\n",$invalid);
print_r($valid);

输出：

{
  "response": {        
    "id": "16" , 
    "event_name": "testing",
    "image": "images/Penguins.jpg",
    "event_date": "2014-12-13",
    "event_time": "02:10",
    "time_interval": "4",
    "location": "",
    "event_type": "Rock" ,
    "detail": "sfdsf fgf ghb\t",
    "delivery": "dggh fghgfh\t\t",
    "status": "1" 
  },
  "bar":{ "test": ""
  }
}

但我们确实建议修复数据库中的数据以及产生这些无效的表单/代码JSON ^^

Answer 2

它在哪些方面无效？如果仅存在双引号，则只使用preg_replace：

$valid = preg_replace('/""/','"',$invalid);

在php中将无效的json转换为有效的json

2 个答案: