| Bug #95140 | mysqld_safe appends --log-syslog-xxx arguments causing super long command line | ||
|---|---|---|---|
| Submitted: | 26 Apr 2019 2:30 | Modified: | 28 Jun 2019 5:19 |
| Reporter: | chiang nigel | Email Updates: | |
| Status: | No Feedback | Impact on me: | |
| Category: | MySQL Server | Severity: | S3 (Non-critical) |
| Version: | 5.7.24 | OS: | Ubuntu |
| Assigned to: | MySQL Verification Team | CPU Architecture: | x86 |
[28 May 2019 5:19]
MySQL Verification Team
Hi, I'm not sure I follow how's this a problem. When does this prevent mysqld from starting? Can you give me example? thanks Bogdan
[29 Jun 2019 1:00]
Bugs System
No feedback was provided for this bug for over a month, so it is being suspended automatically. If you are able to provide the information that was originally requested, please do so and change the status of the bug back to "Open".
Description: mysqld_safe calls `eval_log_error "$cmd"` repeatedly in a while loop when lauching mysqld failed, but `eval_log_error` appends "--log-syslog-xxx" arguments uncondiontionaly, which causing the started mysqld with a super long command line arguments. ```sh eval_log_error () { cmd="$1" case $logging in file) if [ -w / -o "$USER" = "root" ]; then cmd="$cmd > /dev/null 2>&1" else cmd="$cmd >> "`shell_quote_string "$err_log"`" 2>&1" fi ;; syslog) cmd="$cmd --log-syslog=1 --log-syslog-facility=$syslog_facility '--log-syslog-tag=$syslog_tag' > /dev/null 2>&1" ;; both) if [ -w / -o "$USER" = "root" ]; then cmd="$cmd --log-syslog=1 --log-syslog-facility=$syslog_facility '--log-syslog-tag=$syslog_tag' > /dev/null 2>&1" else cmd="$cmd --log-syslog=1 --log-syslog-facility=$syslog_facility '--log-syslog-tag=$syslog_tag' >> "`shell_quote_string "$err_log"`" 2>&1" fi ;; *) echo "Internal program error (non-fatal):" \ " unknown logging method '$logging'" >&2 ;; esac #echo "Running mysqld: [$cmd]" cmd="env MYSQLD_PARENT_PID=$$ $cmd" eval "$cmd" } ``` How to repeat: When mysqld cannot start due to some reason, the mysqld_safe problem will appear.