命令行脚本php不运行

i am trying to build a php script to process data manually to later convert it to a cronjob. this script also get data from mysql and 3rd party soap. when i try to run it from command line i have an error and the script does not run.

我正在尝试构建一个PHP脚本来手动处理数据，以便以后将其转换为cronjob。这个脚本也从mysql和第三方soap获取数据。当我尝试从命令行运行它时，我有一个错误，脚本不运行。

it shows:

表明：

./test.php: line 1: ?php: No such file or directory
Enter a number:
./test.php: line 5: syntax error near unexpected token `('
./test.php: line 5: `$line = trim(fgets(STDIN));'

here's what i have in my script:

这是我脚本中的内容：

 echo 'Enter a number:';
  $line = trim(fgets(STDIN));
  var_dump($line);

i know this script works, what is wrong?

我知道这个剧本有效，有什么不对？

1 个解决方案

#1

You get this error because you execute this script like ./script.php. In order to make sure the PHP script understand and run properly, you have to include this #!/usr/bin/php at the top of your script.

您收到此错误是因为您执行此脚本，如./script.php。为了确保PHP脚本能够理解并正确运行，您必须在脚本的顶部包含＃！/ usr / bin / php。

Example:

例：

#!/usr/bin/php
<?php
echo 'Enter a number:';
$line = trim(fgets(STDIN));
var_dump($line);

if PHP is installed in the /usr/bin folder, if not, you can verify using the locate php command and then use the right path.

如果PHP安装在/ usr / bin文件夹中，如果没有，则可以使用locate php命令进行验证，然后使用正确的路径。

or the other alternative will be

或者另一种选择

php /path/to/script.php

php /path/to/script.php

#1