php - PHP 变量中不返回 MySQL 数据

Question

<?php
require("config.inc.php");

$query = "Select 1 FROM dogs WHERE dog_id = :iddog ";

$query_params = array(':iddog'=> $_POST['iddogPHP2']);

$stmt   = $db->prepare($query);
$result = $stmt->execute($query_params);
$dono2 = $result ->fetchColumn(1);
$raca2 = $result ->fetchColumn(2); 
$sex2 = $result ->fetchColumn(3);
$estado2 = $result ->fetchColumn(4);

$query = "Select * FROM dogs WHERE dispon = 'Sim' AND dono != '$dono2' AND estado = '$estado2' AND raca = '$raca2' AND sex != '$sexo2' ";

try
{
    $stmt   = $db->prepare($query);
    $result = $stmt->execute();
}
catch (PDOException $ex)
{
    $response["success"] = 0;
    $response["message"] = "Database Error!";
    die(json_encode($response));
}

// Finally, we can retrieve all of the found rows into an array using fetchAll
$rows = $stmt->fetchAll();

?>

这是我的代码，问题是变量 dono2、raca2、gender2 没有从数据库中获取值。有什么问题吗？

Answer 1

您需要在第一个查询中使用 SELECT * 而不是 SELECT 1。

$query = "Select * FROM dogs WHERE dog_id = :iddog ";

下面返回一个结果数组

$rows = $stmt->fetchAll();

因此您需要指定要引用的对象。

$dono2 = $rows[0]['dono'];
$raca2 = $rows[0]['raca']; 
$sexo2 = $rows[0]['sex'];
$estado2 = $rows[0]['estado'];