添加 Symfony 应用程序时出现 JSON_ERROR答案

【问题标题】：JSON_ERROR when adding Symfony app添加 Symfony 应用程序时出现 JSON_ERROR
【发布时间】：2017-08-25 17:48:34
【问题描述】：

我刚刚在运行 PHP 7.0.14 的 CentOS 设置上安装了 Symfony。然而，我在尝试运行 symfony new project_name 时遇到以下问题：

    PHP Notice:  Use of undefined constant JSON_ERROR_DEPTH - assumed 'JSON_ERROR_DEPTH' in phar:///usr/local/bin/symfony/vendor/guzzlehttp/guzzle/src/Utils.php on line 134
PHP Notice:  Use of undefined constant JSON_ERROR_STATE_MISMATCH - assumed 'JSON_ERROR_STATE_MISMATCH' in phar:///usr/local/bin/symfony/vendor/guzzlehttp/guzzle/src/Utils.php on line 134
PHP Notice:  Use of undefined constant JSON_ERROR_CTRL_CHAR - assumed 'JSON_ERROR_CTRL_CHAR' in phar:///usr/local/bin/symfony/vendor/guzzlehttp/guzzle/src/Utils.php on line 134
PHP Notice:  Use of undefined constant JSON_ERROR_SYNTAX - assumed 'JSON_ERROR_SYNTAX' in phar:///usr/local/bin/symfony/vendor/guzzlehttp/guzzle/src/Utils.php on line 134
PHP Notice:  Use of undefined constant JSON_ERROR_UTF8 - assumed 'JSON_ERROR_UTF8' in phar:///usr/local/bin/symfony/vendor/guzzlehttp/guzzle/src/Utils.php on line 134
PHP Fatal error:  Uncaught Error: Call to undefined function json_decode() in phar:///usr/local/bin/symfony/vendor/guzzlehttp/guzzle/src/Utils.php:142
Stack trace:
#0 phar:///usr/local/bin/symfony/vendor/guzzlehttp/guzzle/src/Message/Response.php(145): GuzzleHttp\Utils::jsonDecode('{"lts":"2.8.18"...', true, 512, 0)
#1 phar:///usr/local/bin/symfony/src/Symfony/Installer/NewCommand.php(127): GuzzleHttp\Message\Response->json()
#2 phar:///usr/local/bin/symfony/src/Symfony/Installer/NewCommand.php(65): Symfony\Installer\NewCommand->checkSymfonyVersionIsInstallable()
#3 phar:///usr/local/bin/symfony/vendor/symfony/console/Command/Command.php(259): Symfony\Installer\NewCommand->execute(Object(Symfony\Component\Console\Input\ArgvInput), Object(Symfony\Component\Console\Output\ConsoleOutput))
#4 phar:///usr/local/bin/symfony/vendor/symfony/console/Application.php(878): Symfony\Component\Console\Command\Command->run(Object(Symfony\Component\Console\Input\ArgvInput), Object(Symfony\Component\Console\Output\ConsoleOutput))
#5 phar:///us in phar:///usr/local/bin/symfony/vendor/guzzlehttp/guzzle/src/Utils.php on line 142

我已经在网上查了答案，并且已经尝试在 php.ini 中设置 phar.readonly = Off

【问题讨论】：

你是如何安装 php 7 的？似乎您缺少 json 扩展名，或者需要启用它；
你说得对，@hassan，我错过了 JSON 扩展。

标签： php json symfony centos php-7

【解决方案1】：

Hassan 是对的，我错过了 json 扩展。通过运行 sudo yum install php70u-json 修复它

【讨论】：