CODEWITHSUNDEEP
phpmysqlarrays
Chris Nord
Chris Nord

Reputation: 3

mysql array to string conversion

I'm trying to query with a php list, I'm not even sure if it is possible but hopefully you can give some advice/assistance on the matter.
I get a zip list from another function that contains a lot of zip codes, and I want to use that list to narrow down a search.

Notice: Array to string conversion in C:\wamp\www

This is where I get the array to string error . JOIN zip_list zl ON r.zip_code IN ('$zip_list') "

I know it works perfectly with replacing $zip_list with ('2150','2165') for example but the list will be rather long so I can't just put $zip_list[$i].

I've tried some conversion with $trimZip and $test but it always yields the same error.

for($i=0; $i<count($zip_list); $i++){
    array_push($trimZip, "'".$zip_list[$i]."',");
    $i = $i+1;
}
for($i = 0; $i<count($trimZip); $i++){
    $test[$i] = $trimZip[$i];
}

Is there a good way to do this or am I even looking at it the right way ? Better/easier angles are welcome too!

Upvotes: 0

Views: 2739

Answers (2)

Chris Nord
Chris Nord

Reputation: 3

$sql_string = implode(', ', $zip_list);

Implode did the trick, thanks for the help

Upvotes: 0

chris85
chris85

Reputation: 23892

This 

$sql_string = implode(', ', $zip_list);

should do it.

Sample:

$zip_list = array(1, 2);
$sql_string = implode(', ', $zip_list);
echo $sql_string;

Output:

1, 2

Demo: https://eval.in/483839

This presumes you validated the data in $zip_list when assigning it. Variables being used directly in SQL isn't good practice. You should really look into parameterized queries.

I'd do:

$zip_list = array(1, 2);
foreach($zip_list as $param) {
     $placeholders[] = '?';
}
$sql_string = implode(', ', $placeholders);
echo $sql_string;

Then use $sql_string in the prepare and $zip_list in the execute.

Dependent on your driver:

http://php.net/manual/en/mysqli.quickstart.prepared-statements.php
http://php.net/manual/en/pdo.prepared-statements.php

Upvotes: 1

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