check whether table exist

Question

I am trying to check whether a table exists in my database but I am always getting "You are in!" as output. The output of the if statement is not being displayed in the browser.

I appreciate any help.

  $con = new mysqli ( 'localhost', 'root', '', 'bustracker' );
    echo "You are in!";
             if($con->connect_errno){
        die ( "connection problem!".$con->connect_error);
             }

          //check whether route's table exists.
          $results = $con->query("SHOW TABLES LIKE '".$route."'");

           if( ($results->num_rows) == 1){
             echo "Table exist";
           }else{
            echo "table does not exist";
           }

You have an error here: `$conn->connect_error;` need `$con->connect_error`; — mcklayin, Apr 27 '15 at 11:13
If you only get "You are in!" is because your connection doesnt error : all your other code is in the if ( $con->connect_errno ) block. So you never go throught it. — rtome, Apr 27 '15 at 11:14
I have the close bace in the code below and I changed the typo and I am getting nothing — benz, Apr 27 '15 at 11:19
no it is not but I want to check whether does it exist or not and if it exits insert data otherwise create it. — benz, Apr 27 '15 at 11:28
@saty: do I have a problem with $route in the mysql statement? I mean tje way I included it in it — benz, Apr 27 '15 at 11:36
if Bus_4 table not exits then its go into else condition it means your code is working. — Saty, Apr 27 '15 at 11:38

score 3 · Answer 1 · answered Apr 27 '15 at 11:15

3

SELECT 1 FROM tablename LIMIT 1;

If there's no error, table exists.

Or, if you want to be correct, use INFORMATION_SCHEMA.

SELECT * 
FROM information_schema.tables
WHERE table_schema = 'yourdb' 
AND table_name = 'your_table_name'
LIMIT 1;

Alternatively, you can use SHOW TABLES

SHOW TABLES LIKE 'your_table_name';

If there is a row in the resultset, table exists

answered Apr 27 '15 at 11:15

Praveen Dabral

2,449
4
32
46

Yes I found this answer somewhere too and I tried it: http://stackoverflow.com/questions/8829102/mysql-check-if-table-exists-without-using-select-from but no output – benz Apr 27 '15 at 11:22
Yes you answer is right but the error in my code was the if statement which block the code. When I put the code in the else if-statement parts it works – benz Apr 27 '15 at 12:00

Sujeet Kumar · Accepted Answer · 2020-09-15T16:49:47.723

-3

<?php
 
$con = @new mysqli('localhost', 'root', '', 'bustracker');
echo "You are in!";
if ($con->connect_errno) {
    die("connection problem!" . $con->connect_error);
} else {
    $results = $con->query("SHOW TABLES LIKE '" . $route . "'");

    if (($results->num_rows) == 1) {
        echo "Table exist";
    } else {
        echo "table does not exist";
    }
}

edited Sep 15 '20 at 16:49

answered Apr 27 '15 at 11:42

Sujeet Kumar

159
6

Very bad practices in the conception AND the code (avoir @ for ignoring errors...) – David Ansermot Apr 27 '15 at 11:54
if i remove @ then also this code will be execute.@David'mArm'Ansermot – Sujeet Kumar Apr 27 '15 at 12:09
Yeah, but without @, if there is an error, you will be notified. WIth the @, the script will continue with error, it's bad practice. – David Ansermot Apr 27 '15 at 12:12

check whether table exist

2 Answers2