我正在构建一个PHP脚本,将JSON数据提供给另一个脚本。我的脚本将数据构建到一个大型关联数组中,然后使用json_encode输出数据。下面是一个脚本示例:

$data = array('a' => 'apple', 'b' => 'banana', 'c' => 'catnip');
header('Content-type: text/javascript');
echo json_encode($data);

上面的代码产生如下输出:

{"a":"apple","b":"banana","c":"catnip"}

如果你有少量的数据,这是很好的,但我更喜欢这样的东西:

{
    "a": "apple",
    "b": "banana",
    "c": "catnip"
}

有没有办法在PHP中做到这一点,而不需要丑陋的黑客?似乎Facebook的某个人发现了这一点。


当前回答

1 - json_encode($rows,JSON_PRETTY_PRINT);返回带有换行符的美化数据。这对于命令行输入很有帮助,但正如您所发现的那样,在浏览器中看起来不那么漂亮。浏览器将接受换行符作为源(因此,查看页面源确实会显示漂亮的JSON),但它们并不用于在浏览器中格式化输出。浏览器需要HTML。

2 -使用这个功能github

<?php
    /**
     * Formats a JSON string for pretty printing
     *
     * @param string $json The JSON to make pretty
     * @param bool $html Insert nonbreaking spaces and <br />s for tabs and linebreaks
     * @return string The prettified output
     * @author Jay Roberts
     */
    function _format_json($json, $html = false) {
        $tabcount = 0;
        $result = '';
        $inquote = false;
        $ignorenext = false;
        if ($html) {
            $tab = "&nbsp;&nbsp;&nbsp;&nbsp;";
            $newline = "<br/>";
        } else {
            $tab = "\t";
            $newline = "\n";
        }
        for($i = 0; $i < strlen($json); $i++) {
            $char = $json[$i];
            if ($ignorenext) {
                $result .= $char;
                $ignorenext = false;
            } else {
                switch($char) {
                    case '[':
                    case '{':
                        $tabcount++;
                        $result .= $char . $newline . str_repeat($tab, $tabcount);
                        break;
                    case ']':
                    case '}':
                        $tabcount--;
                        $result = trim($result) . $newline . str_repeat($tab, $tabcount) . $char;
                        break;
                    case ',':
                        $result .= $char . $newline . str_repeat($tab, $tabcount);
                        break;
                    case '"':
                        $inquote = !$inquote;
                        $result .= $char;
                        break;
                    case '\\':
                        if ($inquote) $ignorenext = true;
                        $result .= $char;
                        break;
                    default:
                        $result .= $char;
                }
            }
        }
        return $result;
    }

其他回答

对于运行PHP 5.3或之前版本的用户,您可以尝试以下操作:

$pretty_json = "<pre>".print_r(json_decode($json), true)."</pre>";

echo $pretty_json;

这个解决方案生成“非常漂亮”的JSON。这并不完全是OP所要求的,但它让您更好地可视化JSON。

/**
 * takes an object parameter and returns the pretty json format.
 * this is a space saving version that uses 2 spaces instead of the regular 4
 *
 * @param $in
 *
 * @return string
 */
function pretty_json ($in): string
{
  return preg_replace_callback('/^ +/m',
    function (array $matches): string
    {
      return str_repeat(' ', strlen($matches[0]) / 2);
    }, json_encode($in, JSON_PRETTY_PRINT | JSON_HEX_APOS)
  );
}

/**
 * takes a JSON string an adds colours to the keys/values
 * if the string is not JSON then it is returned unaltered.
 *
 * @param string $in
 *
 * @return string
 */

function markup_json (string $in): string
{
  $string  = 'green';
  $number  = 'darkorange';
  $null    = 'magenta';
  $key     = 'red';
  $pattern = '/("(\\\\u[a-zA-Z0-9]{4}|\\\\[^u]|[^\\\\"])*"(\s*:)?|\b(true|false|null)\b|-?\d+(?:\.\d*)?(?:[eE][+\-]?\d+)?)/';
  return preg_replace_callback($pattern,
      function (array $matches) use ($string, $number, $null, $key): string
      {
        $match  = $matches[0];
        $colour = $number;
        if (preg_match('/^"/', $match))
        {
          $colour = preg_match('/:$/', $match)
            ? $key
            : $string;
        }
        elseif ($match === 'null')
        {
          $colour = $null;
        }
        return "<span style='color:{$colour}'>{$match}</span>";
      }, str_replace(['<', '>', '&'], ['&lt;', '&gt;', '&amp;'], $in)
   ) ?? $in;
}

public function test_pretty_json_object ()
{
  $ob       = new \stdClass();
  $ob->test = 'unit-tester';
  $json     = pretty_json($ob);
  $expected = <<<JSON
{
  "test": "unit-tester"
}
JSON;
  $this->assertEquals($expected, $json);
}

public function test_pretty_json_str ()
{
  $ob   = 'unit-tester';
  $json = pretty_json($ob);
  $this->assertEquals("\"$ob\"", $json);
}

public function test_markup_json ()
{
  $json = <<<JSON
[{"name":"abc","id":123,"warnings":[],"errors":null},{"name":"abc"}]
JSON;
  $expected = <<<STR
[
  {
    <span style='color:red'>"name":</span> <span style='color:green'>"abc"</span>,
    <span style='color:red'>"id":</span> <span style='color:darkorange'>123</span>,
    <span style='color:red'>"warnings":</span> [],
    <span style='color:red'>"errors":</span> <span style='color:magenta'>null</span>
  },
  {
    <span style='color:red'>"name":</span> <span style='color:green'>"abc"</span>
  }
]
STR;

  $output = markup_json(pretty_json(json_decode($json)));
  $this->assertEquals($expected,$output);
}

}

PHP 5.4提供了JSON_PRETTY_PRINT选项,用于json_encode()调用。

https://php.net/manual/en/function.json-encode.php

<?php
...
$json_string = json_encode($data, JSON_PRETTY_PRINT);

1 - json_encode($rows,JSON_PRETTY_PRINT);返回带有换行符的美化数据。这对于命令行输入很有帮助,但正如您所发现的那样,在浏览器中看起来不那么漂亮。浏览器将接受换行符作为源(因此,查看页面源确实会显示漂亮的JSON),但它们并不用于在浏览器中格式化输出。浏览器需要HTML。

2 -使用这个功能github

<?php
    /**
     * Formats a JSON string for pretty printing
     *
     * @param string $json The JSON to make pretty
     * @param bool $html Insert nonbreaking spaces and <br />s for tabs and linebreaks
     * @return string The prettified output
     * @author Jay Roberts
     */
    function _format_json($json, $html = false) {
        $tabcount = 0;
        $result = '';
        $inquote = false;
        $ignorenext = false;
        if ($html) {
            $tab = "&nbsp;&nbsp;&nbsp;&nbsp;";
            $newline = "<br/>";
        } else {
            $tab = "\t";
            $newline = "\n";
        }
        for($i = 0; $i < strlen($json); $i++) {
            $char = $json[$i];
            if ($ignorenext) {
                $result .= $char;
                $ignorenext = false;
            } else {
                switch($char) {
                    case '[':
                    case '{':
                        $tabcount++;
                        $result .= $char . $newline . str_repeat($tab, $tabcount);
                        break;
                    case ']':
                    case '}':
                        $tabcount--;
                        $result = trim($result) . $newline . str_repeat($tab, $tabcount) . $char;
                        break;
                    case ',':
                        $result .= $char . $newline . str_repeat($tab, $tabcount);
                        break;
                    case '"':
                        $inquote = !$inquote;
                        $result .= $char;
                        break;
                    case '\\':
                        if ($inquote) $ignorenext = true;
                        $result .= $char;
                        break;
                    default:
                        $result .= $char;
                }
            }
        }
        return $result;
    }

我也有同样的问题。

不管怎样,我只是在这里使用了json格式代码:

http://recursive-design.com/blog/2008/03/11/format-json-with-php/

能满足我的需要。

还有一个更加维护的版本:https://github.com/GerHobbelt/nicejson-php