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命令行登录mysql提示Segmentation fault解决办法
时间:2022-06-30 21:09:18 编辑:袖梨 来源:一聚教程网
使用时报错如下
[root@localhost bin]# ./mysql -uroot -p Enter password: Welcome to the MySQL monitor. Commands end with ; or g. Your MySQL connection id is 8 Server version: 5.6.24 Source distribution Copyright (c) 2000, 2015, Oracle and/or its affiliates. All rights reserved. Oracle is a registered trademark of Oracle Corporation and/or its affiliates. Other names may be trademarks of their respective owners.
Segmentation fault
mysql版本是5.6.24
服务器是CentOS7.0
==========解决方法1============
在源码包里,编辑文件 cmd-line-utils/libedit/terminal.c
把terminal_set方法中的 char buf[TC_BUFSIZE]; 这一行注释,再把 area = buf;改为 area = NULL;
更改后如下
protected int terminal_set(EditLine *el, const char *term) { int i; /*char buf[TC_BUFSIZE];*/ char *area; const struct termcapstr *t; sigset_t oset, nset; int lins, cols; --phpfensi.com (void) sigemptyset(&nset); (void) sigaddset(&nset, SIGWINCH); (void) sigprocmask(SIG_BLOCK, &nset, &oset); area = NULL;
然后重新编译即可
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