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[Solved] If statement in bash

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Top Forums Shell Programming and Scripting [Solved] If statement in bash
# 1  
Old 01-10-2014
[Solved] If statement in bash
I have the following code in bash, however "set red frmt" is not displayed.

Code:
  echo "iarg_rd = $iarg_rd"

  iarg_rd="2"
  if [ ! -z "$iarg_rd" ]; then
    echo "Hello World"
  fi

  if [ ! -z "$iarg_colr" ]; then
    frmt="${gap}${!frmt_titl_yl}"

  elif [ ! -z "$iarg_bk" ]; then
    frmt="${gap}${!frmt_titl_bk}"

  elif [ ! -z "$iarg_rd" ]; then
    echo "set red frmt"
    frmt="${gap}${!frmt_titl_rd}"

  elif [ ! -z "$iarg_gn" ]; then
    frmt="${gap}${!frmt_titl_gn}"

  elif [ ! -z "$iarg_yl" ]; then
    frmt="${gap}${!frmt_titl_yl}"

  elif [ ! -z "$iarg_bl" ]; then
    frmt="${gap}${!frmt_titl_bl}"

  elif [ ! -z "$iarg_mg" ]; then
    frmt="${gap}${!frmt_titl_mg}"

  elif [ ! -z "$iarg_cy" ]; then
    frmt="${gap}${!frmt_titl_cy}"

  elif [ ! -z "$iarg_wh" ]; then
    frmt="${gap}${!frmt_titl_wh}"

  else
    frmt="${gap}${!frmt_titl}"

  fi

  echo "frmt = $frmt"

# 2  
Old 01-10-2014
So what does it display then?
# 3  
Old 01-10-2014
Problem fixed.
"$iarg_colr" happened to be 1. Also needed to remove ! and just use the variable name.
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