php mysqli Error: Column 'id' in where clause is ambiguous Statement

Question

I have two table for category name and post. Now i need to show category name for each post like this :

$joinedcontent = Access::FETCH("SELECT * FROM " . CONTENT . 
" LEFT JOIN " . POST . "  ON category = id WHERE  id = ?", $row['id']);
$ccats= $joinedcontent[0]['name'];

In result i see this error:

Error: Column 'id' in where clause is ambiguous Statement: SELECT * FROM cms_content LEFT JOIN cms_posts ON type = id WHERE id = ? Arguments: Array ( [۰] => 30 )

how do fix this error?

possible duplicate of [MySQL: "Column 'column\_name' in where clause is ambiguous"](http://stackoverflow.com/questions/337034/mysql-column-column-name-in-where-clause-is-ambiguous) — Sougata Bose, Aug 20 '15 at 10:08
both tables have column with name id, try to use table alias and use the alias to avoid the ambiguity — abdullacm, Aug 20 '15 at 10:10
Please mark the best answer, if any, as accepted in order to help future visitors. — Ahmad, Sep 10 '15 at 17:49

Ahmad · Accepted Answer · 2015-08-20T10:47:39.070

1

This is an ambiguity for the database that the "id" column specified in where clause, belongs to which table. CONTENT or POST? You have to specify the name of the table before the column name to identify it. For example:

$joinedcontent = Access::FETCH("SELECT * FROM " . CONTENT . 
" LEFT JOIN " . POST . "  ON category =  " . POST . ".id WHERE  " . POST . ".id = ?", $row['id']);
$ccats= $joinedcontent[0]['name'];

edited Aug 20 '15 at 10:47

answered Aug 20 '15 at 10:12

Ahmad

5,551
8
41
57

Yes, I missed one of them. The edited one should work. – Ahmad Aug 20 '15 at 10:48

score 0 · Answer 2 · answered Aug 20 '15 at 10:12

0

Might be same column available in other table. so please use table.column to avoid conflict column (ambigity)

answered Aug 20 '15 at 10:12

pmr

14
1

Seems to be comment more – Narendrasingh Sisodia Aug 20 '15 at 10:24

php mysqli Error: Column 'id' in where clause is ambiguous Statement

2 Answers2