PHP in_array function not working with mysqli_fetch_array

Question

If I do not use MySQL, the PHP in_array function works as expected. In this example, the if statement would execute and display the array contains the keyword blue because the in_array function is able to find keyword blue in the array.

<?php
$keyword = "blue";
$array = array("red", "green", "blue", "yellow");

if (in_array($keyword, $array)) {echo "the array contains keyword $keyword";}
else {echo "the array does not contain keyword $keyword";}
?>

in_array fails to do the same when selecting red, green, blue, yellow from MySQL. In this example, if I echo $array, red green blue yellow is displayed in the Web browser, so I know blue is in the $array variable. However, the else statement executes and displays the array does not contain keyword blue.

<?php
$keyword = "blue";

$con = new mysqli('domain','username','password','database');
$sql = "select * from colors";
$sql_query = $con->query($sql);

while ($row = mysqli_fetch_array($sql_query)) {
  $array = $row['colors'];

  if (in_array($keyword, $array)) {echo "the array contains keyword $keyword";}
  else {echo "the array does not contain keyword $keyword";}
}
?>

If anyone has any tips or suggestions here, I certainly would appreciate it.

Thanks for the tip! I am able to successfully get the "if" statement to trigger with array_key_exists when not using MySQL. However, array_key_exists is still causing the "else" statement to trigger when getting the $array using MySQL. Thanks for the suggestion here! :) — JeremyCanfield
– JeremyCanfield, Commented Jan 17, 2016 at 1:07

cFreed · Accepted Answer · 2016-01-17 01:14:09Z

2

In your SQL version you get a record where a field named colors should exist.
But this field contains a string (maybe a comma-separated list, something like blue,red,...), not an array.

So when you execute in_array($keyword, $array) you fire a PHP error (but likely you have not currently error_report set to show them).

What you must do is to transform the content you got in colors field into an array.
In the case it is like evoked above:

$array = explode(',', $row['colors']);

Or whatever appropriate depending on what is exactly the colors field structure.

edited Jan 17, 2016 at 1:14

answered Jan 17, 2016 at 1:09

cFreed

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1 Comment

JeremyCanfield Over a year ago

oh I saw references to explode before but I didn't recognize this as the way to handle this situation. Wow, I really appreciate this answer!

KJS · Accepted Answer · 2023-10-11 18:59:57Z

0

I used this answer. And it still works perfectly.

If you have an array with only values you should use in_array function as this example:

$array = array("A", "B", "C");
in_array($string, $array) ? "found" : "not found";

If you have an array with keys you should use isset function as this example:

$array = array("A" => "Result A", "B" => "Result B", "C" => "Result C");
isset($array[$string]) ? "found" : "not found";
or like:
if(isset($array[$string])) { do your stuff... }

answered Oct 11, 2023 at 18:59

KJS

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