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I took Jeet's answer and adapted it to Windows (thanks to this answer):

FOR /F %x IN ('"git rev-list --all"') DO @git grep <regex> %x > out.txt

Note that for me, for some reason, the actual commit that deleted this regex did not appear in the output of the command, but rather one commit prior to it.

I took Jeet's answer and adapted it to Windows (thanks to this answer):

FOR /F %x IN ('"git rev-list --all"') DO @git grep <regex> %x > out.txt

Note that for me, for some reason, the actual commit that deleted this regex did not appear in the output of the command, but rather one commit prior to it.

I took @Jeet's answer and adpated it to Windows (thanks to this answer):

FOR /F %x IN ('"git rev-list --all"') DO @git grep <regex> %x > out.txt

Note that for me, for some reason, the actual commit that deleted this regex did not appear in the output of the command, but rather one commit prior to it.

I took Jeet's answer and adapted it to Windows (thanks to this answer):

FOR /F %x IN ('"git rev-list --all"') DO @git grep <regex> %x > out.txt

Note that for me, for some reason, the actual commit that deleted this regex did not appear in the output of the command, but rather one commit prior to it.

I took @Jeet's answer and adpated it to Windows (thanks to this answer):

FOR /F %x IN ('"git rev-list --all"') DO @git grep <regex> %x > out.txt

Note that for me, for some reason, the actual commit that deleted this regex did not appear in the output of the command, but rather one commit prior to it.

I took @Jeet's answer and adpated it to Windows (thanks to this answer):

FOR /F %x IN ('"git rev-list --all"') DO @git grep <regex> %x > out.txt

Note that for me, for some reason, the actual commit that deleted this regex did not appear in the output of the command, but rather one commit prior to it.