Question

我使用docker-compose创建一个使用Nginx和PostgreSQL操作的zabbix容器，但它生成以下错误Parse错误：

语法错误，意外＆＃39 ;;＆＃39;在/etc/zabbix/web/zabbix.conf.php上第24行。

我不明白这一点，因为它在几分钟前工作正常。

搬运工-compose.yaml

version: '3.1'
services:
postgres:
image: postgres
restart: always
environment:
  POSTGRES_USER: zabbix
  POSTGRES_PASSWORD: zabbix
  POSTGRES_DB: zabbix
zabbix-server:
image: zabbix/zabbix-server-pgsql
restart: always
environment:
  DB_SERVER_HOST: postgres
  POSTGRES_USER: zabbix
  POSTGRES_PASSWORD: zabbix
  POSTGRES_DB: zabbix
depends_on:
  - postgres
zabbix-web:
image: zabbix/zabbix-web-nginx-pgsql
restart: always
environment:
  ZBX_SERVER_HOST: zabbix-server
  DB_SERVER_HOST: postgres
  POSTGRES_USER: zabbix
  POSTGRES_PASSWORD: zabbix
  POSTGRES_DB: zabbix
depends_on:
  - postgres
  - zabbix-server
ports:
  - 8081:80

Answer 1

检查您的zabbix.con.php文件是否具有此形状。您可以在zabbix Web容器中找到此文件。

<?php
// Zabbix GUI configuration file.
global $DB, $HISTORY;
$DB['TYPE']     = 'POSTGRESQL';
$DB['SERVER']   = 'postgres';
$DB['PORT']     = '5432';
$DB['DATABASE'] = 'zabbix';
$DB['USER']     = 'zabbix';
$DB['PASSWORD'] = 'zabbix';
// Schema name. Used for IBM DB2 and PostgreSQL.
$DB['SCHEMA'] = '';
$ZBX_SERVER      = 'zabbix-server';
$ZBX_SERVER_PORT = '10051';
$ZBX_SERVER_NAME = 'Zabbix docker';
$IMAGE_FORMAT_DEFAULT   = IMAGE_FORMAT_PNG;
// Elasticsearch url (can be string if same url is used for all types).
$HISTORY['url']   = '';
// Value types stored in Elasticsearch.
$HISTORY['types'] = '';

通常会在代码中的某处出现一个等于后跟分号的错误，从而产生错误。

解析错误：语法错误，意外＆＃39 ;;＆＃39;在第24行的/etc/zabbix/web/zabbix.conf.php中

1 个答案: