replace links in output
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i have a question about the replace_in_output function.
i wanna do something like that:
But when i do this i get following error-message:
Parse error: syntax error, unexpected '/' in ......\xampp\htdocs\fusiontest\includes\output_handling_include.php(96) : eval()'d code on line 1
Merged on Apr 24 2012 at 21:07:19:
I fixed it by doing this:
is there a replace function which automatically escapes all these signs?
i wanna do something like that:
Code Download source
replace_in_output("<!-- TEST_FLAG -->", "<a href='../index.php'>Test</a>");
But when i do this i get following error-message:
Parse error: syntax error, unexpected '/' in ......\xampp\htdocs\fusiontest\includes\output_handling_include.php(96) : eval()'d code on line 1
Merged on Apr 24 2012 at 21:07:19:
I fixed it by doing this:
Code Download source
$repl = "<a href='../index.php'>Test</a>";
$repl = str_replace("'","\'",$repl);
replace_in_output("<!-- TEST_FLAG -->",$repl);
is there a replace function which automatically escapes all these signs?
Edited by mawe4585 on 24-04-2012 20:07,
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