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
格式化JSON数据的最佳方式是这样的!
header('Content-type: application/json; charset=UTF-8');
echo json_encode($response, JSON_PRETTY_PRINT | JSON_UNESCAPED_UNICODE | JSON_UNESCAPED_SLASHES);
将$response替换为需要转换为JSON的数据
其他回答
递归解的经典例子。这是我的:
class JsonFormatter {
public static function prettyPrint(&$j, $indentor = "\t", $indent = "") {
$inString = $escaped = false;
$result = $indent;
if(is_string($j)) {
$bak = $j;
$j = str_split(trim($j, '"'));
}
while(count($j)) {
$c = array_shift($j);
if(false !== strpos("{[,]}", $c)) {
if($inString) {
$result .= $c;
} else if($c == '{' || $c == '[') {
$result .= $c."\n";
$result .= self::prettyPrint($j, $indentor, $indentor.$indent);
$result .= $indent.array_shift($j);
} else if($c == '}' || $c == ']') {
array_unshift($j, $c);
$result .= "\n";
return $result;
} else {
$result .= $c."\n".$indent;
}
} else {
$result .= $c;
$c == '"' && !$escaped && $inString = !$inString;
$escaped = $c == '\\' ? !$escaped : false;
}
}
$j = $bak;
return $result;
}
}
用法:
php > require 'JsonFormatter.php';
php > $a = array('foo' => 1, 'bar' => 'This "is" bar', 'baz' => array('a' => 1, 'b' => 2, 'c' => '"3"'));
php > print_r($a);
Array
(
[foo] => 1
[bar] => This "is" bar
[baz] => Array
(
[a] => 1
[b] => 2
[c] => "3"
)
)
php > echo JsonFormatter::prettyPrint(json_encode($a));
{
"foo":1,
"bar":"This \"is\" bar",
"baz":{
"a":1,
"b":2,
"c":"\"3\""
}
}
干杯
我意识到这个问题是问如何编码一个关联数组到一个漂亮的JSON格式的字符串，所以这没有直接回答这个问题，但如果你有一个字符串，已经在JSON格式，你可以很简单地通过解码和重新编码它(需要PHP >= 5.4):
$json = json_encode(json_decode($json), JSON_PRETTY_PRINT);
例子:
header('Content-Type: application/json');
$json_ugly = '{"a":1,"b":2,"c":3,"d":4,"e":5}';
$json_pretty = json_encode(json_decode($json_ugly), JSON_PRETTY_PRINT);
echo $json_pretty;
这个输出:
{
"a": 1,
"b": 2,
"c": 3,
"d": 4,
"e": 5
}
我从Composer: https://github.com/composer/composer/blob/master/src/Composer/Json/JsonFile.php和nicejson: https://github.com/GerHobbelt/nicejson-php/blob/master/nicejson.php中获取代码 Composer代码很好，因为它从5.3到5.4的更新很流畅，但它只编码对象，而nicejson需要json字符串，所以我合并了它们。代码可以用来格式化json字符串和/或编码对象，我目前在Drupal模块中使用它。
if (!defined('JSON_UNESCAPED_SLASHES'))
define('JSON_UNESCAPED_SLASHES', 64);
if (!defined('JSON_PRETTY_PRINT'))
define('JSON_PRETTY_PRINT', 128);
if (!defined('JSON_UNESCAPED_UNICODE'))
define('JSON_UNESCAPED_UNICODE', 256);
function _json_encode($data, $options = 448)
{
if (version_compare(PHP_VERSION, '5.4', '>='))
{
return json_encode($data, $options);
}
return _json_format(json_encode($data), $options);
}
function _pretty_print_json($json)
{
return _json_format($json, JSON_PRETTY_PRINT);
}
function _json_format($json, $options = 448)
{
$prettyPrint = (bool) ($options & JSON_PRETTY_PRINT);
$unescapeUnicode = (bool) ($options & JSON_UNESCAPED_UNICODE);
$unescapeSlashes = (bool) ($options & JSON_UNESCAPED_SLASHES);
if (!$prettyPrint && !$unescapeUnicode && !$unescapeSlashes)
{
return $json;
}
$result = '';
$pos = 0;
$strLen = strlen($json);
$indentStr = ' ';
$newLine = "\n";
$outOfQuotes = true;
$buffer = '';
$noescape = true;
for ($i = 0; $i < $strLen; $i++)
{
// Grab the next character in the string
$char = substr($json, $i, 1);
// Are we inside a quoted string?
if ('"' === $char && $noescape)
{
$outOfQuotes = !$outOfQuotes;
}
if (!$outOfQuotes)
{
$buffer .= $char;
$noescape = '\\' === $char ? !$noescape : true;
continue;
}
elseif ('' !== $buffer)
{
if ($unescapeSlashes)
{
$buffer = str_replace('\\/', '/', $buffer);
}
if ($unescapeUnicode && function_exists('mb_convert_encoding'))
{
// http://stackoverflow.com/questions/2934563/how-to-decode-unicode-escape-sequences-like-u00ed-to-proper-utf-8-encoded-cha
$buffer = preg_replace_callback('/\\\\u([0-9a-f]{4})/i',
function ($match)
{
return mb_convert_encoding(pack('H*', $match[1]), 'UTF-8', 'UCS-2BE');
}, $buffer);
}
$result .= $buffer . $char;
$buffer = '';
continue;
}
elseif(false !== strpos(" \t\r\n", $char))
{
continue;
}
if (':' === $char)
{
// Add a space after the : character
$char .= ' ';
}
elseif (('}' === $char || ']' === $char))
{
$pos--;
$prevChar = substr($json, $i - 1, 1);
if ('{' !== $prevChar && '[' !== $prevChar)
{
// If this character is the end of an element,
// output a new line and indent the next line
$result .= $newLine;
for ($j = 0; $j < $pos; $j++)
{
$result .= $indentStr;
}
}
else
{
// Collapse empty {} and []
$result = rtrim($result) . "\n\n" . $indentStr;
}
}
$result .= $char;
// If the last character was the beginning of an element,
// output a new line and indent the next line
if (',' === $char || '{' === $char || '[' === $char)
{
$result .= $newLine;
if ('{' === $char || '[' === $char)
{
$pos++;
}
for ($j = 0; $j < $pos; $j++)
{
$result .= $indentStr;
}
}
}
// If buffer not empty after formating we have an unclosed quote
if (strlen($buffer) > 0)
{
//json is incorrectly formatted
$result = false;
}
return $result;
}
如果你正在使用MVC
尝试在您的控制器中执行此操作
public function getLatestUsers() {
header('Content-Type: application/json');
echo $this->model->getLatestUsers(); // this returns json_encode($somedata, JSON_PRETTY_PRINT)
}
然后如果你调用/getLatestUsers，你会得到一个漂亮的JSON输出;)
我通常会用这些简单的语句。
如果你已经有一个JSON字符串，你可以简单地使用echo()和print_r()的组合。 不要忘记将print_r()的第二个参数传递给true，以便它返回值而不是打印它:
echo('<pre>' . print_r($json, true) . '</pre>');
或者你可以使用die()，这对调试很方便:
die('<pre>' . print_r($json, true) . '</pre>');
如果您有一个数组，您需要在之前将其转换为JSON字符串。 请确保使用JSON_PRETTY_PRINT标志设置json_encode()的第二个参数，这样您的JSON将被正确呈现:
echo('<pre>' . print_r(json_encode($array, JSON_PRETTY_PRINT), true) . '</pre>');
或者用于调试:
die('<pre>' . print_r(json_encode($array, JSON_PRETTY_PRINT), true) . '</pre>');
