How to avoid inserting data twice in MySQL using PHP and mysqli

Question

I'm trying to catch error messages returned from a mysql server on insert failure. The below method works fine when fetching data, but on insert, the 'if' statement below inserts the data a second time. How can I re-write this to catch error messages without inserting the data again.

    $mysqli = new mysqli("localhost", "user", "pass", "database");

    $query_storeReturnedData = "INSERT INTO `eventStore` (table_key, event_type, event_date,) VALUES(NULL, 'Unix Event', '2010-08-24 12:00:00')";

    $mysqli->query($query_storeReturnedData);

    if(!$mysqli->query($query_storeReturnedData))
    {
     printf("Errormessage: %s\n", mysqli_error($mysqli));
    }

codaddict · Accepted Answer · 2010-08-24 04:00:19Z

2

Because you have two calls to $mysqli->query you are seeing double insertions.

You can remove the first call to $mysqli->query

// no need of this.
// $mysqli->query($query_storeReturnedData);

// make just one call and also do err checking.
if(!$mysqli->query($query_storeReturnedData)) {
   printf("Errormessage: %s\n", mysqli_error($mysqli));
}

Alternatively you can collect the return value of $mysqli->query and use it in error checking:

// run the query and collect the return value.
$return_value = $mysqli->query($query_storeReturnedData);

// if return value is false..query failed...print err msg.
if(!$return_value)) {
   printf("Errormessage: %s\n", mysqli_error($mysqli));
}

answered Aug 24, 2010 at 4:00

codaddict

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1 Comment

JMC Over a year ago

Your alternate solution explains it. Should have been evaluating against the return value not the function call. Going to store this under 'duh!'. Thanks

Wolph · Accepted Answer · 2010-08-24 04:00:44Z

2

How about this?

$mysqli = new mysqli("localhost", "user", "pass", "database");

$query_storeReturnedData = "INSERT INTO `eventStore` (table_key, event_type, event_date,) VALUES(NULL, 'Unix Event', '2010-08-24 12:00:00')";

if(!$mysqli->query($query_storeReturnedData))
{
 printf("Errormessage: %s\n", mysqli_error($mysqli));
}

answered Aug 24, 2010 at 4:00

Wolph

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