Question

如何迭代关联数组并使用其值创建多个记录。

问题似乎是正确配置bind_param的变量。

array(2) {
  [0]=>
  object(stdClass)#4 (3) {
    ["number"]=>int(1)
    ["detail"]=>string(3) "one"
  }
 [1]=>
 object(stdClass)#5 (3) {
   ["number"]=>int(2)
   ["detail"]=>string(3) "two"
 }

}

PHP和MYSQL

foreach ($myArray as $key => $value){
    $one = $value->number;
    $two = $value->detail;

    $sqlQuery = "INSERT INTO my_tab (number, detail) VALUES (?,?)";

    if($statement = $conexion->prepare($sqlQuery)){
       $statement->bind_param("is", $one , $two);
       $statement->execute();
    }else{
    ...
    }
}

my_tab中的预期结果：

number   detail
  1        one
  2        two

Answer 1

我认为没有任何理由不这样做：

$sqlQuery = "INSERT INTO my_tab (number, detail) VALUES (?,?)"; // no need to include this in loop - it's just a static string

if ($conexion) { // make sure this is defined
    foreach ($myArray as $key => $value){
        $one = (int)$value->number; // enforce this type as int
        $two = $value->detail;

        $statement = $conexion->prepare($sqlQuery));
        if ($statement) { 
            $statement->bind_param("is", $one , $two);
            $statement->execute();
        }else{
            ...
        }
    }
}

如何迭代关联数组并使用其值创建多个记录

1 个答案: