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Full Discussion: Passing variable from PHP to bash script
Top Forums Web Development Passing variable from PHP to bash script Post 303009059 by kieranfoley on Friday 8th of December 2017 11:14:29 AM
Old 12-08-2017
This the bash script (dpa_report.sh) that creates the REST -API report. I have taken out the actual username and password for the purposes of this post. If I run this script from the command line and pass it the hostname it works. For example

Code:
./dpa_report hostname

works. The hostname is set to $1 in the script.

But when dpa_report.sh is called from the php script I get the permission errors.


Code:
#!/bin/bash

export HOME=/var/www/html/scripts/REST-API
export TEMP_DIR=/var/www/html/scripts/REST-API/Temp

curl  -u "username:password" http://seieadpa01:9004/apollo-api/nodes/?query=name%3d$1 > $TEMP_DIR/test_file.txt

HOST_ID=`cat $TEMP_DIR/test_file.txt | grep "<id>"`

cat $HOME/host_query_input_file.out | awk -v ID="$HOST_ID" '{if ($1~/\<id>/) $1=ID ;print}' > $TEMP_DIR/host_query_input_file.out

curl -u username:password -X POST -H "Content-Type:application/vnd.emc.apollo-v1+xml" --data "@/var/www/html/scripts/REST-API/host_query_input_file.out" http://seieadpa01:9004/dpa-api/report -o $TEMP_DIR/test_xml.out

LINK=$(grep link $TEMP_DIR/test_xml.out)
LINK="${LINK#*>}"
LINK="${LINK%<*}"

export LINK

sleep 5


curl -u username:password $LINK > /var/www/html/REPORTS/$1.html

 

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link

LINK(1) 							   User Commands							   LINK(1)

NAME

       link - call the link function to create a link to a file

SYNOPSIS

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       link OPTION

DESCRIPTION

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       --help display this help and exit

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AUTHOR

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REPORTING BUGS

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SEE ALSO

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	      info coreutils 'link invocation'

       should give you access to the complete manual.

GNU coreutils 8.12.197-032bb					  September 2011							   LINK(1)
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