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在Symfony 2中创建json文件_Json_Symfony - Fatal编程技术网

在Symfony 2中创建json文件

在Symfony 2中创建json文件,json,symfony,Json,Symfony,嗨,我已经创建了这样的json对象: {"ask":[{"rate":"2110.00000000","amount":"2.00000000","price":"4220.00000000","degree":0.2},{"rate":"1800.00000000","amount":"5.00000000","price":"9000.00000000","degree":0.5},{"rate":"20.00000000","amount":"8.00000000","price":"16

嗨,我已经创建了这样的json对象:

{"ask":[{"rate":"2110.00000000","amount":"2.00000000","price":"4220.00000000","degree":0.2},{"rate":"1800.00000000","amount":"5.00000000","price":"9000.00000000","degree":0.5},{"rate":"20.00000000","amount":"8.00000000","price":"160.00000000","degree":0.8}],"bid":[{"rate":"1700.00000000","amount":"0.50000000","price":"1700.00000000","degree":0.5}]} 
我想创建一个文件并将json放入其中,我如何才能做到这一点?

假设您有包含json字符串的
$json

$fs = new \Symfony\Component\Filesystem\Filesystem();

try {
    $fs->dumpFile('path/to/my/file.json', $json);
}
catch(IOException $e) {
}
或者您可以使用PHP函数,例如

假设您有包含json字符串的
$json

$fs = new \Symfony\Component\Filesystem\Filesystem();

try {
    $fs->dumpFile('path/to/my/file.json', $json);
}
catch(IOException $e) {
}

或者可以使用PHP函数,例如

如何“创建”上述JSON?考虑把它保存到一个文件中吗?它是从数据库中恢复过来的,我使用JSONY-EnCODE()函数。我需要将它保存到文件中,因为这样可以缩短Ajax的访问时间。您是如何“创建”上述JSON的?考虑把它保存到一个文件中吗?它是从数据库中恢复过来的,我使用JSONY-EnCODE()函数。我需要将其保存到文件中,因为这将缩短从Ajax的访问时间,路径应该是例如“AcmeDemoBunle/Data/my.json?Nop。这是文件系统上的路径。例如“c:\\Windows”(绝对路径)或“tmp”(相对路径)。您可以使用类似$This->get('kernel')->getRootDir();要获取Symfony2目录,路径应为“AcmeDemoBunle/Data/my.json?Nop”。这是文件系统上的路径。例如“c:\\Windows”(绝对路径)或“tmp”(相对路径)。您可以使用类似$this->get('kernel')->getRootDir()的思想;要获取Symfony2目录