convert mysql result to json with correct types

Question

I know how to get a mysql-row and convert it to json:

$row = mysqli_fetch_assoc(mysqli_query($db, \"SELECT * FROM table WHERE id=1\"));
echo json_encode($row         


        

Answer 1

A MySQLi OO version based on @axiac's answer, that produces a JSON array ($jsnAll) containing all records. In this code snippet, the method FixSQLType is called to fix a row. Note, it should be wrapped in a try{}catch{} block and "objMySQLi" has already been instantiated:

         $lcAllRows = array();

         // Make an SQL SELECT statement 
         $SQL = "SELECT * FROM $lcName WHERE $lcWhere";

         // Run the query
         $this->sqlResult =  $this->objMySQLi->query($SQL);

         // Fetch the result
         while( $row = $this->sqlResult->fetch_assoc()){
             $lcCount = count($lcAllRows) ;

             // Call to fix, row 
             $fixedRow = $this->FixSQLType($row);
             $lcAllRows[$lcCount]= $fixedRow;
         }

         $jsnAll = json_encode($lcAllRows);

The FixSQLType method. This is almost identical to @axiac's answer, except for the call to $this->sqlResult->fetch_field_direct($i). "fetch_field" seemed to get itself lost, using "fetch_field_direct" overcame that problem.

    private function FixSQLType($pRow){
    // FROM https://stackoverflow.com/a/28261996/7571029

    // Fix the types    
    $fixed = array();
    $i = 0;
    foreach ($pRow as $key => $value) {
        $info = $this->sqlResult->fetch_field_direct($i);
        $i++;

        if (in_array($info->type, array(
                MYSQLI_TYPE_TINY, MYSQLI_TYPE_SHORT, MYSQLI_TYPE_INT24,    
                MYSQLI_TYPE_LONG, MYSQLI_TYPE_LONGLONG,
                MYSQLI_TYPE_DECIMAL, 
                MYSQLI_TYPE_FLOAT, MYSQLI_TYPE_DOUBLE
        ))) {
            $fixed[$key] = 0 + $value;
        } else {
            $fixed[$key] = $value;
        }

    }

    return $fixed;
}