CSV find and replace with quotation marks with PHP

Question

Trying to figure out what I am dong wrong here but I am trying to search row 6 and replace the phrase "Plenty of stock available!" (with the quotation marks as it is in a CSV) with the number "10".

It works no issues when then with the 'lowstock' text, but not the "Plenty of stock available!". I have added the \ before it as to have it ignore the quotation mark (and view it as text).

 //changes stock status
    $row = fgetcsv($file);
    if (strtolower($row[6]) == '\"Plenty of stock available!\"' || strtolower($row[6]) == 'lowstock') {
    $row[6] = '10';
    } else if (strtolower($row[6]) == 'nostock') {
    $row[6] = '2';

strtolower($row[6]) will never match a P, did you mean p instead? — Tigger
– Tigger, Commented Nov 8, 2017 at 4:58
strtolower($row[6]) == '\"Plenty of stock available!\"' need to be strtolower($row[6]) == strtolower('"Plenty of stock available!"') — Death-is-the-real-truth
– Death-is-the-real-truth, Commented Nov 8, 2017 at 5:04
I don't get why you wouldn't just use a lower case 'p' in the string. — TimBrownlaw
– TimBrownlaw, Commented Nov 8, 2017 at 5:08

Tigger · Accepted Answer · 2017-11-08 05:09:55Z

1

I would try the following:

$row = fgetcsv($file);
// Don't modify the original data (unless you need to) and 
// remove duplicate calls to strtolower()
$r6 = strtolower($row[6]);
// Only test against lower case letters, " should not be needed
if ($r6 == 'plenty of stock available!' || $r6 == 'lowstock') {
    // perform correction / change
    $row[6] = '10';
} else if ($r6 == 'nostock') {
    // perform correction / change
    $row[6] = '2';
}

answered Nov 8, 2017 at 5:09

Tigger

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2 Comments

Phil Over a year ago

@LukeAppel if that's the case, then your row[6] value never contained quote characters

Tigger Over a year ago

@LukeAppel - Cool, glad to have helped.

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