Jordi_E

[quote author=wildteen88 link=topic=100916.msg399309#msg399309 date=1153244586] Change this: [code]return ($this->num_rows() == 1); [/code] to this: [code]return (($this->num_rows() == 1) ? true : false);[/code] The change this: [code]if (!$this->BOX_exists( $BOX_id ))[/code] to this: [code]if ($this->BOX_exists( $BOX_id ) === true)[/code] [/quote] Thanks a lot for your ideas. I tried it, but I still have the same problem. On my explanation I forgot to mention that these functions are inside of a class which inherits from another class called DB, which is implementing the query and the Database connection

[quote author=redarrow link=topic=100916.msg399285#msg399285 date=1153242888] try it this way but only chaged this  ($this->num_rows() > 0); ps. i had a go my self got the same result two inserts dont no why theo. i read the code and basickly its saying if theres one insert but the code says if there 1 insert with another one. can not work it out theo sorry. ps i do get the inpression becouse your pulling one function into another that why there a double entry. function BOX_exists( $BOX_id ) {     $sql = "SELECT IDBOX FROM box ".            "WHERE IDBOX='".$BOX_id."';";     $this->query($sql);     return ($this->num_rows() > 0); } function insert_BOX( $BOX_id, $Description) {     if (!$this->BOX_exists( $BOX_id ))     {         $sql =  "INSERT INTO box (IDBOX, Description )".                 "VALUES ('$BOX_id', '$Description');";         $this->query( $sql );         return true;     }     else     {         return false;     } } try this ok. [code] function insert_BOX( $BOX_id, $Description) {       $sql = "SELECT IDBOX FROM box ".            "WHERE IDBOX='".$BOX_id."';";     $this->query($sql);     return ($this->num_rows() > 0);   if (!$this->BOX_exists( $BOX_id ))     { $sql =  "INSERT INTO box (IDBOX, Description )".                 "VALUES ('$BOX_id', '$Description');";         $this->query( $sql );         return true;     }     else     {         return false;     } } [/code] [/quote] Thanks for your answer. I tried what you suggest, but it still behaves bad.

[quote author=businessman332211@hotmail.com link=topic=100916.msg399230#msg399230 date=1153237012] Try This [code]function BOX_exists( $BOX_id ) {     $sql = "SELECT IDBOX FROM box ".            "WHERE IDBOX='".$BOX_id."';";     $this->query($sql);     return ($this->num_rows() == 1); } function insert_BOX( $BOX_id, $Description) {     if (BOX_exists( $BOX_id ))     {         $sql =  "INSERT INTO box (IDBOX, Description )".                 "VALUES ('$BOX_id', '$Description');";         $this->query( $sql );         return true;     }     else     {         return false;     } }[/code] That should make it stop doing that. [/quote] Thanks a lot for your answer. But obviously I use  if (!BOX_exists( $BOX_id ))  prior inserting a new BOX. If it exists in the Database, I wish not to insert it. So, changing from (!BOX_exists( $BOX_id )) to (BOX_exists( $BOX_id )) is not the solution. Best regards.

Hello everybody. I'm newbie in PHP. I am running PHP version 5.1.2 together with Apache version 2.0.52 and mySQL version 4.1 on a Windows XP system. Consider these functions: [color=green]function BOX_exists( $BOX_id ) {     $sql = "SELECT IDBOX FROM box ".           "WHERE IDBOX='".$BOX_id."';";     $this->query($sql);     return ($this->num_rows() == 1); } function insert_BOX( $BOX_id, $Description) {     if (!$this->BOX_exists( $BOX_id ))     {         $sql =  "INSERT INTO box (IDBOX, Description )".                 "VALUES ('$BOX_id', '$Description');";         $this->query( $sql );         return true;     }     else     {         return false;     } }[/color] The problem I have is that calling to insert_BOX function, always inserts correctly data on mySQL, but the return value from insert_BOX is sometimes false (not always, sometimes is true). As can be seen, this should be impossible, since data is correctly inserted. I only call insert_BOX once, but what would happen in case insert_BOX is executed twice for some reason? The second call would return false. I feel this is the problem I'm having. I don't know how to handle this problem. I spent many hours without success. I will appreciate any idea or help. Thanks in advance.