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Editorial Team
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Asked: 2026-05-29T08:00:43+00:00

I’m working on a code that provide basic information about the place you live

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I’m working on a code that provide basic information about the place you live in, in the Netherlands.

There for i need the ‘id’ from a zip code that can vary between zip A (1234) of zip B (1236). Those begin and end numbers are already in the database. But I failed to get the id, because i’m getting an error with the SQL syntax.

I have tried to do the following:

SELECT `id` WHERE `9291` BETWEEN 'zipA' AND 'zipB'
SELECT `id` WHERE `zipA` > '9291' OR `zipB` < '9291'
SELECT `id` WHERE `zipA` > '9291' AND `zipB` < '9291'

What could i do to make this work? I’m just a beginner with MYSQL and php, i’m currently working with CodeIgniter 2.1.0

Thank you very much!

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1 Answer

  1. Editorial Team
    SELECT `id` WHERE `9291` BETWEEN 'zipA' AND 'zipB'
SELECT `id` WHERE `9291` BETWEEN 'zipA' AND 'zipB'
SELECT `id` WHERE `zipA` > '9291' AND `zipB` < '9291'

    A couple of things:

    • You cannot compare an integer value to a char/varchar value. It
      looks to me like you are trying to compare the column 9291 to the
      string values ‘zipA’ AND ‘zipB’.
    • Be careful with your backticks / single quotes.
    • Where is your FROM statement? You are selecting column id FROM ?

    Could you please try something like this:

    SELECT
   id
FROM
   zip_code_table
WHERE
   zip BETWEEN 9000 AND 9100
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