在PHP

Question

在PHP中，我试图构建一个需要在一个表中检查ID的表单，如果它存在，它应该在另一个表中创建一个记录。到目前为止，这是有效的，但我遇到的问题是，如果我检查的ID不存在，我试图处理案件。如果它不存在，我想创建另一个。但每次尝试时，我都会在获取结果时从服务器获得500个错误。

基本上我做了以下功能

function trySQL($con, $query, $params) {
$stmt = $con->prepare($query); 
$result = $stmt->execute($params); 
$id = $stmt->insert_id;
return array($stmt,$result,$id);
}

我通过我的PHP代码多次调用此函数但是当我再调用它一次并尝试获取结果时它会中断。

   $custINSquery = "
        INSERT INTO custs ( 
            FirstName, 
            LastName, 
            EmailAddress, 
            PhoneNumber 
        ) VALUES ( 
            :FirstName, 
            :LastName, 
            :EmailAddress, 
            :PhoneNumber
        )
    ";

    $createJob = " 
        INSERT INTO jobs ( 
            custs_id,
            StAddress, 
            State, 
            ZipCode,
            MoistureLocation,
            status_id
        ) VALUES ( 
            :custs_id, 
            :StAddress, 
            :State, 
            :ZipCode,
            :IssueDesc,
            :status_id
        ) 
    "; 

    $custSELquery = "SELECT id, FirstName, LastName, EmailAddress FROM custs WHERE FirstName = :FirstName AND LastName = :LastName AND EmailAddress = :EmailAddress";

    $custSELquery_params = array( 
        ':FirstName' => $_POST['FirstName'],
        ':LastName' => $_POST['LastName'],
        ':EmailAddress' => $_POST['EmailAddress']
    ); 

    $checkcust = trySQL($db, $custSELquery, $custSELquery_params);

    $row = $checkcust[0]->fetch(); 

    if(!$row){
        $custINSquery_params = array( 
            ':FirstName' => $_POST['FirstName'],
            ':LastName' => $_POST['LastName'],
            ':EmailAddress' => $_POST['EmailAddress'],
            ':PhoneNumber' => $_POST['PhoneNumber']
        ); 

        $custins = trySQL($db, $custINSquery, $custINSquery_params);
        $custsel = trySQL($db, $custSELquery, $custSELquery_params);

        $custs_id = $custsel[0]->fetch();
        if ($custs_id != null) {
            $createJobParam = array( 
            ':custs_id' => $custs_id,
            ':StAddress' => $_POST['StAddress'],
            ':State' => $_POST['State'],
            ':ZipCode' => $_POST['ZipCode'],
            ':IssueDesc' => $_POST['MoistureLocation'],
            ':status_id' => $_POST['status_id']
            ); 
            $jobins = trySQL($db, $createJob, $createJobParam);
            $jobres = $jobins[0]->fetch();
            die("um...");
            if ($jobres) {
                # code...
                die("looks like I made it");
            }
        }

    } else {
        $createJobParam = array( 
        ':custs_id' => $row['id'],
        ':StAddress' => $_POST['StAddress'],
        ':State' => $_POST['State'],
        ':ZipCode' => $_POST['ZipCode'],
        ':IssueDesc' => $_POST['MoistureLocation'],
        ':status_id' => $_POST['status_id']
        ); 
        $data['success'] = true;
        $data['message'] = 'Success!';            
    }

附加说明：当我查看php文档时，他们说我可以使用inserted_id的东西来获取我之前插入的ID但是当我尝试它时它只是给了我这个集合的空值起来。

任何帮助都将不胜感激。

谢谢！