带有php的PreparedStatement select未按预期工作

我有一个疑问：

SELECT id, result, ip_address, added_date
FROM results
WHERE course_id = ( 
SELECT id
FROM courses
WHERE course =  'informatica'
AND macro_course_id = ( 
SELECT id
FROM macro_courses
WHERE macro_course =  'scienze-matematiche-fisiche-e-naturali'
AND organization_id = ( 
SELECT id
FROM organizations
WHERE organization =  'universita-degli-studi-di-torino'
AND city_id = ( 
SELECT id
FROM cities
WHERE city =  'torino'
AND region_id = ( 
SELECT id
FROM regions
WHERE region =  'piemonte' ))))) ORDER BY id DESC

我用这个代码来做一个准备好的声明

public function getResults($region, $city, $organization, $macro_course, $course) { //works
    //added_date=datetime : YYYY-MM-DD HH:mm:ss
    echo "SELECT id, result, ip_address, added_date
    FROM results
    WHERE course_id = ( 
    SELECT id
    FROM courses
    WHERE course =  '$course'
    AND macro_course_id = ( 
    SELECT id
    FROM macro_courses
    WHERE macro_course =  '$macro_course'
    AND organization_id = ( 
    SELECT id
    FROM organizations
    WHERE organization =  '$organization'
    AND city_id = ( 
    SELECT id
    FROM cities
    WHERE city =  '$city'
    AND region_id = ( 
    SELECT id
    FROM regions
    WHERE region =  '$region' ))))) ORDER BY id DESC"; //just for me to know what query is being executed 


    if ($stmt = $this->mysqli->prepare(("
    SELECT id, result, ip_address, added_date
    FROM results
    WHERE course_id = ( 
    SELECT id
    FROM courses
    WHERE course =  ?
    AND macro_course_id = ( 
    SELECT id
    FROM macro_courses
    WHERE macro_course =  ?
    AND organization_id = ( 
    SELECT id
    FROM organizations
    WHERE organization =  ?
    AND city_id = ( 
    SELECT id
    FROM cities
    WHERE city =  ?
    AND region_id = ( 
    SELECT id
    FROM regions
    WHERE region =  ? ))))) ORDER BY id DESC

  "))) {
        $return = array();
        $stmt->bind_param('sssss', $course, $macro_course, $organization, $city, $region);
        $stmt->execute();
        if ($stmt->fetch()) {
            $i = 0;
            while ($row = $stmt->fetch()) {
                print_r($row);//this is never reached
                continue;
                $s = new Result($row['result'], $row['added_date'], $row['id']);
                $return[$i] = $s;
                $i+=1;
            }
        }
    }
    return $return;
}

问题是该函数返回0行和0个错误（使用

$this->mysqli->error

进行检查），似乎$row=$stmt->fetch（）总是false

但是，如果我复制并在PHPMyAdmin上执行我在函数顶部得到的输出，我会看到

Showing lines 0 - 0 ( 1 total, Query time 0.0003 sec)

---

因此，查询返回一行，但php无法捕获它。我错过了什么？如何修复此问题？

因为您在此处使用了两次

$stmt-fetch（）

if ($stmt->fetch()) {
        $i = 0;
        while ($row = $stmt->fetch()) {

删除
如果（$stmt->fetch（））
条件，它将按预期工作

编辑

来自文档

请注意，在调用$stmt->fetch（）之前，应用程序必须绑定所有列。

在像这样调用
$stmt->fetch（）
之前，必须先绑定结果

/* bind result variables */
$stmt->bind_result($name, $code);

您可以通过手动学习正确的mysqli语法来修复它，或者更好的方法是使用PDO。您是否检查了
query
是否在没有服务器端的情况下工作？查询本身看起来让我眼花缭乱。你不能直接用连接进行一次查询吗？他已经在使用pdo了…我想说，如果需要的话，重组他的表并使用连接，一次查询6次是很糟糕的…@Ranjith阅读了所有问题：
但是，如果我复制并在PHPMyAdmin上执行我在函数顶部得到的输出，我会看到“显示0-0行”（总共1个，查询时间0.0003秒）“
当然可以，否则PMA会显示有一些错误。有时最好关闭一两个过于本地化的问题，而不是回答它。只是为了它。