I want to run this script bash
#!/bin/bash
# file name: myscript.sh
PROJECT_DIR=$1
mkdir $PROJECT_DIR
mkdir $PROJECT_DIR/PolySkills
mkdir $PROJECT_DIR/PolySkills/cr
mkdir $PROJECT_DIR/PolySkills/cr/corrections
mkdir $PROJECT_DIR/PolySkills/cr/corrections/jpg
mkdir $PROJECT_DIR/PolySkills/cr/corrections/pdf
mkdir $PROJECT_DIR/PolySkills/cr/diagnostic
mkdir $PROJECT_DIR/PolySkills/cr/zooms
mkdir $PROJECT_DIR/PolySkills/data
mkdir $PROJECT_DIR/PolySkills/exports
mkdir $PROJECT_DIR/PolySkills/scans
mkdir $PROJECT_DIR/PolySkills/copies
cd $PROJECT_DIR/PolySkills
cp ~/file.tex $PROJECT_DIR/PolySkills
For the variable "PROJECT_DIR" that represents a path in which I will create folders, I want to retrieve its value from a php variable. I looked at some examples on the internet and I tried one but it does not work. This is what i used :
chdir('~/');
$pathtofile = "~/ExportEval/".$NomUE."/".$NomOcc."/".$Numtudiant;
$directory=$pathtofile."/AMC_Project";
$output = exec("./myscript $directory");
Knowing that the script file "myscript " exists in home "~ /"
thank you for your help :)
I update my question : I found what is the problem but I don't see what the solution as the variable $NomUE is composed of a sentence separated by spaces it considers that $ 1 is only the first of this sentence if I change $ 1 by $ 2 it takes the second word of that same sentence! I don't understand why it does not take $pathtofile as path !
You need to always escape variables before sending them to the command line. Also, ~
has no meaning here and there's no reason to use chdir()
. Just use a fully qualified pathname instead:
<?php
$pathtofile = "/home/someuser/ExportEval/$NomUE/$NomOcc/$Numtudiant";
$directory = escapeshellarg("$pathtofile/AMC_Project");
$output = exec("/home/someuser/myscript $directory");
You may have problems running a script in a home directory, and reading from that directory. Best is to put the script in a proper location such as /usr/local/bin/
.
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