MySQL_Insert_Id

[Xtremer360]

How can you debug to find out why it's not getting the mysql_insert_id number? I have it echoed all the queries and all with all the correct values from my form but the only problem is that its not getting the insert id number of the id.

 

$query1 = "INSERT INTO `efed_bio` (charactername,username,posername,style_id,gender,status_id,division_id,alignment_id,sortorder) VALUES ('".$charactername."','".$username."','".$posername."','".$style."','".$gender."','".$status."','".$division."','".$alignment."','".$sort."')";
            mysql_query($query1);
            $query1_id = mysql_insert_id(); 
            echo $query1;
            echo $query1_id;
            $query2 = "INSERT INTO `efed_bio_allies` (bio_id) VALUES (".$query1_id.")";
            mysql_query($query2); 
            echo $query2;
            $query3 = "INSERT INTO `efed_bio_rivals` (bio_id) VALUES (".$query1_id.")";
            mysql_query($query3);  
            echo $query3;
            $query5 = "INSERT INTO `efed_bio_singles` (bio_id) VALUES (".$query1_id.")";
            mysql_query($query5);  
            echo $query5;

[JAY6390]

Is the field set to auto_increment?

[chintansshah]

Please find detail function inputs and outputs on php.net website

 

http://in3.php.net/manual/en/function.mysql-insert-id.php

[litebearer]

perhaps...

 

 

$query1 = "INSERT INTO efed_bio (charactername, username, posername, style_id, gender, status_id, division_id, alignment_id, sortorder) VALUES ('$charactername', '$username', '$posername', '$style', '$gender', '$status', '$division', '$alignment', '$sort')";$result1 = mysql_query($query1);$query1_id = mysql_insert_id(); echo $query1;echo $query1_id;$query2 = "INSERT INTO efed_bio_allies (bio_id) VALUES ('$query1_id')";$result2 = mysql_query($query2); echo $query2;$query3 = "INSERT INTO efed_bio_rivals (bio_id) VALUES ('$query1_id')";$result3 = mysql_query($query3); echo $query3;$query5 = "INSERT INTO efed_bio_singles (bio_id) VALUES ('$query1_id')";$result5 = mysql_query($query5); echo $query5;

 

[Pawn]

How can you debug code that doesn't bother to check if what its doing is working? Echo'ing SQL isn't going to help. As chintansshah suggested, read the manual entry and check for unexpected return values.

 

 

Return Values

 

The ID generated for an AUTO_INCREMENT column by the previous query on success, 0 if the previous query does not generate an AUTO_INCREMENT value, or FALSE  if no MySQL connection was established.

 

 

$sql = "INSERT INTO..."; // snipif(!$query = mysql_query($sql)) {    // mysql_query returned FALSE    echo "Error: a query failed on line ".__LINE__.". ".mysql_error();    exit;}if(!$insert_id = mysql_insert_id()) {    // mysql_insert_id returned FALSE    echo "Error: no MySQL connection was established.";    exit;} elseif($insert_id == 0) {    // mysql_insert_id returned 0    echo "Error: the previous query did not generate an AUTO_INCREMENT value.";    exit;}

 

[Xtremer360]

Is the field set to auto_increment?

 

 

Yes

[schilly]

What does

echo $query1_id;

output?

 

Does the row get inserted into efed_bio?

[Xtremer360]

yes

[Xtremer360]

I finally figured out stupidly that there was a field I was declaring in my query that I didn't have in my db table.