# Stumped with Apache 2.2, Mod_fcgid, suexec, php-fastcgi



## LordOfLA (Mar 8, 2010)

My suexec log is giving me the following error:


```
[2010-03-08 19:57:46]: uid: (1000/[censored]) gid: (1000/[censored]) cmd: php.fcgi
[2010-03-08 19:57:46]: target uid/gid (1000/1000) mismatch with directory (1000/1000) or program (0/34359738368)
```
The command in question php.fcgi is as follows:


```
-rwx------  1 censored  censored    38 Mar  8 19:49 php.fcgi
```

Where is suexec getting 0/34359738368 from?

I've been banging away at this for a few hours whittling down configuration and permission errors/mistakes and can't figure this last one out. I'm getting the feeling this is the last one.

This is on the following:


```
FreeBSD [censored] 8.0-RELEASE-p2 FreeBSD 8.0-RELEASE-p2 #0: Sat Mar  6 11:45:51 UTC 2010     root@[censored]:/usr/obj/usr/src/sys/GENERIC  amd64
```

Many thanks in advance for any suggestions.


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## LordOfLA (Mar 8, 2010)

Hmmm.. I seem to have fixed it and I have no idea how.. since the last thing I did was run the following commands:


```
find . -type f -name "*.php" -exec chmod 750 {} \;
find . -type d -exec chmod 755 {} \;
```
I can only assume I missed a permission setting somewhere.

Thanks for looking anyway!


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## LordOfLA (Mar 8, 2010)

that would be `find . -type f -name "*.php" -exec chmod 755 {} \;` even...


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## DutchDaemon (Mar 8, 2010)

Please use proper formatting tags.


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