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I'm having a nightmare installing ZF with the Zend Tool. The manual says I have to install some folders in the includes path and two files in the system executable path. But I don't know what this means.

 

I'm using a Linux hosted shared server with access to Shell and tried a few things ("which php"which points to a system folder I don't have permission to edit and editting a bash file) but still can't the zend tool running.

 

Please someone help  :confused:

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PHP's include path is defined within the php.ini. It's generally set to somewhere like /usr/share/php, which means you would put the Zend directory found within library in /usr/share/php

 

As for your systems execution path, this can be found easily by typing....

 

echo $PATH

 

In a terminal.

 

Generally, I create a local bin directory and place this on my $PATH, then any custom scripts I need on my path go in /home/<username>/bin

 

You can add to your current $PATH by editing your .bashrc file found within your home directory. I line like:

 

PATH=$PATH:$HOME/bin; export PATH

 

Will do what I described above.

Thank you for your help thorpe

 

I've gone back to changing the bash files and it loads ok now (was loading zf.bat before) but only from inside the bin directory for the zf.sh file. I've tried creating a link so I can access the file from any directory like this:

 

ln -s public_html/zend_training/bin/zf.sh public_html/zend_training/bin/zf

 

but i get command not found. Any idea why this is?

 

 

Is public_html/zend_training/bin on your path?

 

I checked it with echo $PATH as you said.

 

I can access zf inside the -s public_html/zend_training/bin folder if I do do this inside the same folder:

 

ln -s zf.sh zf

 

If I access outside the directory, however, it says "Could not open input file: ./zf.php"

 

On the otherhand, if I create the link at the route level to the same bin folder, I get -jailshell: Could: command not found. Using this:

 

ln -s public_html/zend_training/bin/zf.sh public_html/zend_training/bin/zf

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