asked on

Parse error: syntax error, unexpected '->'

Im converting some of my code from old style procedure to OOP in order to tidy up the code a bit better, however Im a little confused on my error saying "Parse error: syntax error, unexpected '->'"

The error Im getting is on the line:-

        global $mysqli->query("select `ticket`.*, `job`.*, `joblocation`.* from `ticket` left join `job` on `job`.`serialNumber` = `ticket`.`jobSerialNumber` left join `joblocation` on `job`.`locationID` = `joblocation`.`locationID` where ((" . $this->buildWhereCompany() . ") and (" . $this->buildWhereSite() . ") and (" . $this->buildWherejobManufacturer() . "));");

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So the variable $mysqli is declared outside the class, in another class, but it seems to be raising the parse error in the building of the SQL statment.

My code which replicates the issue:-

$mysqli = new mysqli($dbHost,$dbUsername,$dbPassword,$dbName);

class claTicketTable {
  function __construct() {
        global $mysqli->query("select `ticket`.*, `job`.*, `joblocation`.* from `ticket` left join `job` on `job`.`serialNumber` = `ticket`.`jobSerialNumber` left join `joblocation` on `job`.`locationID` = `joblocation`.`locationID` where ((" . $this->buildWhereCompany() . ") and (" . $this->buildWhereSite() . ") and (" . $this->buildWherejobManufacturer() . "));");
  }

  private function buildWhereCompany() {
    $sqlWhere = "(`joblocation`.`company` like \"%\")";
    return $sqlWhere;
  }
  private function buildWhereSite() {
    $sqlWhere = "(`joblocation`.`site` like \"%\")";
    return $sqlWhere;
  }
  private function buildWherejobManufacturer() {
    $sqlWhere = "(`joblocation`.`manufacturer` like \"%\")";
    return $sqlWhere;
  }
  function drawTable() {
  }
}

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Can anyone see what Im doing wrong?????

Dave Baldwin

I don't know if you can use 'global' there but you are definitely not assigning the results of $mysqli->query to a variable. Without assigning it to a variable you can't use the results.

http://us2.php.net/mysqli_query

tonelm54

ASKER

Sorry, dont know how I missed that from my example code, it is there....

If I use:-

$resTable = $mysqli->query("select uuid();");

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I get "Notice: Undefined variable: mysqli"

If I use:-

$resTable = global $mysqli->query("select uuid();");

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I get "Parse error: syntax error, unexpected 'global'"

tonelm54

ASKER

I think my issue is with getting access to the variable $mysqli which is created outside my class, but dont know how (or even it if it is possible) to either get access the the variable $mysqli created externally, or how to pass the reference of the variable through as there is no point in creating a new instance.

Thank you

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Chris Stanyon

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tonelm54

ASKER

Thank you :-)

Parse error: syntax error, unexpected '-&gt;'

Parse error: syntax error, unexpected '->'