Linux下php5.4启动脚本的简单示例

这篇文章主要为大家详细介绍了Linux下php5.4启动脚本的简单示例，具有一定的参考价值，可以用来参考一下。

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废话不多说，直接上步骤
1、修改php-fpm.conf配置文件
修改/usr/local/php/etc/php-fpm.conf(当然这个跟你配置的php路径相关)配置文件

启动pid=run/php-fpm.pid

代码如下：

; Pid file
; Note: the default prefix is /usr/local/php/var
; Default Value: none
pid = run/php-fpm.pid

Linux下php5.4启动脚本

启动error_log=log/php-fpm.log

代码如下：

; Error log file
; If it's set to "syslog", log is sent to syslogd instead of being written
; in a local file.
; Note: the default prefix is /usr/local/php/var
; Default Value: log/php-fpm.log
error_log = log/php-fpm.log

Linux下php5.4启动脚本

2、创建/etc/init.d/php文件

代码如下：

root@aliyun:~# cat /etc/init.d/php
#!/bin/bash
#Author:iranw
#E-mail:wang_wenguan#yeah.net
#Website:http://www.phpno.com
 
# pid file
pid="/usr/local/php/var/run/php-fpm.pid"
# php-fpm path&file
phpfpm="/usr/local/php/sbin/php-fpm"
 
param=$1
 
phpnum=`ps -ef | grep php-fpm | wc -l`
 
if [ "$param" = "stop" ]; then
    if [ "$phpnum" = "1" ]; then
        echo "Php process does not exist."
        exit 1
    fi
    kill `cat $pid`
    echo "Close php processes. OK"
    exit 1
fi
 
if [ "$param" = "start" ]; then
    if [ "$phpnum" != "1" ]; then
        echo "Php process exist."
        exit 1
    fi
    $phpfpm
    echo "Open php processes. OK"
    exit 1
fi
 
if [ "$param" = "restart" ]; then
    if [ "$phpnum" = "1" ]; then
        echo "Php process does not exist. Stop Failed."
    else
        kill `cat $pid`
        for i in `ps -ef | grep php-fpm | grep -v grep | awk {'print $2'}`
        do
            kill -9 $i
            echo "Kill $i successed."
        done
        echo "Close php processes. OK"
    fi
    $phpfpm
    if [ "$phpnum" != "1" ]; then
        echo "Open php processes. OK"
    else
        echo "Open php processes. Failed."
    fi
    exit 1
fi

Linux下php5.4启动脚本

3、修改权限

代码如下：

#chmod +x /etc/init.d/php

Linux下php5.4启动脚本

4、执行结果

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