mysql table creation failed

Question

well i followed the answer of a stack overflow question but the table wasn't created in phpmyadmin and neither an erro was shown.here's my code:

$query = "SELECT ID FROM ap_reg";
$result = mysqli_query($con, $query);

if(empty($result)) {
            $query = "CREATE TABLE ap_reg (
                      ID int(11) AUTO_INCREMENT,
                      TITLE varchar(255) ,
                      NAME varchar(255) ,
                      EMAIL varchar(255) ,
                      PHONE varchar(255) ,
                      COUNTRY varchar(255) ,
                      POSTAL_CODE varchar(255) ,
                      TYPE varchar(255) ,
                      APARTMENT_TYPE varchar(255) ,
                      SOURCE varchar(255) ,
                      FINANCE varchar(255) ,
                      PRICE_RANGE varchar(255) ,
                      PURCHASE_WINDOW varchar(255) ,
                      COMMENT TEXT ,
                      DATE varchar(255) ,
                      TIME varchar(255) ,
                      PRIMARY KEY  (ID)
                      )";
            $result = mysqli_query($con, $query);
}

and her's the link to that question

stack overflow question

you didnt get error because you have not used mysqli_error($con) — krishna, Feb 11 '14 at 13:36
yes it was inside if loop and i have messed while pasting it here. i am editing it now. please answer if you can. — buet_baba, Feb 11 '14 at 13:37
No sure but you want to create a table if your query doesn't give you an answer. In that case maybe this query is faster. `CREATE TABLE IF NOT EXISTS ap_reg ... ` — Mr. Radical, Feb 11 '14 at 13:38
@Tularis TYPE can be used without tickmarks and ID is not a keyword. for keywords list check here http://dev.mysql.com/doc/refman/5.5/en/reserved-words.html — krishna, Feb 11 '14 at 13:44
change `$result = mysqli_query($con, $query)` to `$result = mysqli_query($con, $query) or die(mysqli_error($con));` to find if any error is there in query execution — krishna, Feb 11 '14 at 13:45

Mr. Radical · Accepted Answer · 2014-02-11T18:28:42.007

Try this:

$query = "CREATE TABLE IF NOT EXISTS ap_reg ( ID int(11) AUTO_INCREMENT, 
    TITLE varchar(255) , 
    NAME varchar(255) , 
    EMAIL varchar(255) , 
    PHONE varchar(255) , 
    COUNTRY varchar(255) , 
    POSTAL_CODE varchar(255) , 
    TYPE varchar(255) , 
    APARTMENT_TYPE varchar(255) , 
    SOURCE varchar(255) , 
    FINANCE varchar(255) , 
    PRICE_RANGE varchar(255) , 
    PURCHASE_WINDOW varchar(255) , 
    COMMENT TEXT , 
    DATE varchar(255) , 
    TIME varchar(255) , 
    PRIMARY KEY (ID) )"; 

mysqli_query($con, $query) or die(mysqli_error());

This will eliminate the need to test if you get results in you first query and will direct create a table if this doesn't already exist.

However, if you insist you do not want to use IF NOT EXISTS can also try this:

$query = "SELECT ID FROM ap_reg";

$result = mysqli_query($con, $query); 

if(empty($result)) { 
    $query = "CREATE TABLE ap_reg ( ID int(11) AUTO_INCREMENT, 
    TITLE varchar(255) , 
    NAME varchar(255) , 
    EMAIL varchar(255) , 
    PHONE varchar(255) , 
    COUNTRY varchar(255) , 
    POSTAL_CODE varchar(255) , 
    TYPE varchar(255) , 
    APARTMENT_TYPE varchar(255) , 
    SOURCE varchar(255) , 
    FINANCE varchar(255) , 
    PRICE_RANGE varchar(255) , PURCHASE_WINDOW varchar(255) ,
     COMMENT TEXT , DATE varchar(255) , 
    TIME varchar(255) , 
    PRIMARY KEY (ID) )"; 

mysqli_query($con, $query) or die(mysqli_error ());
     }

mysql table creation failed

1 Answers1