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Devin Gray
39,261 Pointsenhancing a simple php application => adding search controller view => escaping output review
Here's the quiz question: "A visitor to our site has just performed a search for chocolate, which has returned no results. Change this code to display the search term in the search box again, making sure to protect the page against malicious code that might have been entered."
I have no idea on what to do, I've looked at other forum topics and I'm almost positive my answer is right. Is there something that I'm missing?
<input type="text" name="s" value="<?php if(isset($search_term)) { echo htmlspecialchars($search_term); } ?>">
I've also tried replacing the $search_term variable with $_GET and $s and they all say the same thing:
'To place text inside a text input field, you should give it a value
attribute.'
Thanks.
Saransh Kalia
516 Points <input type="text" value="<?php if (!$products){ echo htmlspecialchars($search_term); }?>" name="s">

Devin Gray
39,261 PointsOh my goodness! Thank you! Worked like a charm! :)
Saransh Kalia
516 PointsNo worries dear ! :)
2 Answers

Kang-Kyu Lee
52,045 PointsThis might be... "performed a search for chocolate, which has returned no results" part could be considered as a condition. However for me, "isset($search_term)" worked also.

alexismaillard
2,023 PointsYes, I also preferred use "isset($search_term)" function. It worked well:
<input type="text" name="s" value="<?php if (isset($search_term)) {echo htmlspecialchars($search_term); }?>">

Niccolò Mineo
8,687 PointsMay someone explain why the condition is "!$products", please?
Thank you
Saransh Kalia
516 PointsSaransh Kalia
516 PointsPlease use this
<input type="text" value="<?php if (!$products){ echo htmlspecialchars($search_term); }?>" name="s">