I am using following php code.
$i = "1TEN";
$val=exec('cat '$i'.dssp -n | grep " ACC " | grep "[0-9]\+" -o | head -n 1');
echo $val;
It doesn't give any error or any output. However, following code works well:
$val=exec('cat 1TEN.dssp -n | grep " ACC " | grep "[0-9]\+" -o | head -n 1');
echo $val;
Can anyone help ?
Similarly,
$line=exec("tail $i.dssp -n $diff | awk -F" " -v var=$pos '{if ($2==var) print FNR}");
You are receiving this mistake because you are inserting variable in a string in a wrong way – you should use double quotes " to specify your string and escape " in a text with \ so it won't be interpreted as the end of a string:
$val = exec("cat {$i}.dssp -n | grep \" ACC \" | grep \"[0-9]\+\" -o | head -n 1");
or use use double quotes " to specify your string and replace " in text with single quotes':
$val = exec("cat {$i}.dssp -n | grep ' ACC ' | grep '[0-9]\+' -o | head -n 1");
If you want to use single quotes ' you can concatenate you string like it is shown below :
$val = exec('cat ' . $i . ' .dssp -n | grep " ACC " | grep "[0-9]\+" -o | head -n 1');
if ({$pos}==var) print FNR for shell or for php to evaluate?\\{ and \\}Use double quotes on the outside to allow variable interpolation, and use single quotes on the inside.
$val=exec("cat $i.dssp -n | grep ' ACC ' | grep '[0-9]\+' -o | head -n 1");