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Posted by Forest Liu on 03/19/05 02:56
it shuold be:
$SQL_Query = "INSERT INTO ".$this->Company_Referrer_Table." ( Record_ID
, Record_Data ) VALUES ( '', '".$this->Record_Data."')";
just use the operator '.' to connect two string. I think php could
only identify simple car direcctly as $foo to be its value in a string
contain it. The reference to a class data member would not work well
in a string. So you should use the operator '.' to connect them.
On Sat, 19 Mar 2005 02:43:15 +0200, Ahmed Abdel-Aliem <me2resh@gmail.com> wrote:
> hi
>
> i am new to classes and i have a problem with a class i wrote
>
> here is the class
>
> -----------------------------------------------------------------------
> class Company {
>
> var $Company_Name;
> var $Company_Referrer_Table;
> var $Company_Visitor_Table;
> var $Company_ID;
> var $Message;
> var $File;
> var $FileName;
> var $Record_Data;
>
> function CompanyInfo (){
>
> $SQL_Query = mysql_query("SELECT * FROM company_list WHERE
> Company_ID = $this->Company_ID");
> $Record = mysql_fetch_array($SQL_Query);
>
> $this->Company_ID = $Record['Company_ID'];
> $this->Company_Name = $Record['Company_Name'];
> $this->Company_Referrer_Table = $Record['Company_Referrer_Table'];
> $this->Company_Visitor_Table = $Record['Company_Visitor_Table'];
>
> return $this->Company_Referrer_Table;
> return $this->Company_Visitor_Table;
>
> }
>
> function GenerateUniqueID(){
>
> return md5(uniqid(mt_rand(),TRUE));
>
> }
>
> function UploadFile($File) {
>
> $this->File = $File;
> $dst_file_name = $this->GenerateUniqueID();
> $arr = split("\.",$this->File['name']);
> $this->File['name'] = $dst_file_name;
> $this->File['name'] .= ".".$arr[count($arr)-1];
> $dest_dir = 'temp';
> $dest = $dest_dir . '/' .$this->File['name'];
> $r = move_uploaded_file($this->File['tmp_name'], $dest);
> $this->FileName = $this->File['name'];
> chmod("temp/".$this->FileName, 777);
>
> }
>
> function TakeData($File, $Company_ID, $Destination) {
>
> $this->Company_ID = $Company_ID;
>
> $this->CompanyInfo();
>
> if ($Destination == "Referrer"){
>
> $SQL_Query = "INSERT INTO $this->Company_Referrer_Table ( Record_ID
> , Record_Data ) VALUES ( '', '$this->Record_Data')";
>
> }elseif ($Destination == "Visitor"){
>
> $SQL_Query = "INSERT INTO $this->Company_Visitor_Table ( Record_ID
> , Record_Data ) VALUES ( '', '$this->Record_Data')";
>
> }
>
> $this->UploadFile($File);
>
> $fh = fopen("temp/".$this->FileName, "r");
> while (! feof($fh)) : $this->Record_Data = fgets($fh, 4096);
> mysql_query($SQL_Query)or die("Error In Importing Records To The Database");
> endwhile;
> fclose($fh);
>
> $this->Message = "Date Imported Successfully";
> return $this->Message;
>
> }
>
> }
>
> -------------------------------------------------------------------------------
>
> the problem occurs when i call the function TakeDate
>
> i use this code to call it
>
> $Company = new Company();
> $File =& $HTTP_POST_FILES['FileToUpload'];
> $Company->TakeData($File, $Company_ID, $Destination);
> $Message = $Company->Message;
>
> the error i get is "Error In Importing Records To The Database" which
> i defined as the error of mysql in the function.
>
> i checked everything and found that the problem happens because i
> can't retireve the values of
> $this->Company_Referrer_Table and $this->Company_Visitor_Table by
> calling the function CompanyInfo
> from the function TakeData
>
> can anyone advise me with the right way to call the CompanyInfo
> function from the TakeData and get the values correctly?
>
> --
> Ahmed Abdel-Aliem
> Web Developer
> www.ApexScript.com
> 0101108551
> registered Linux user number 382789
>
> --
> PHP General Mailing List (http://www.php.net/)
> To unsubscribe, visit: http://www.php.net/unsub.php
>
>
--
Sincerely,
Forest Liu(刘云赟)
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