Echo失败错误

Question

下面的代码应该连接到应用程序，并将数据插入到已从用户输入的数据库中。但是，当我运行代码时出现一个奇怪的错误：“解析错误：语法错误，意外的'在...中插入''。对我来说，这个错误告诉我，回声'插入成功'，是以某种方式导致代码崩溃，我误解了错误是什么意思？有人能告诉我如何修复代码吗？

<?php 
    require "conn.php"; 

    $patient_name = $_POST["patient_name"]; 
    $doctor_name = $_POST["doctor_name"]; 
    $check_in_date = $_POST["check_in_date"]; 
    $room_number = $_POST["room_number"]; 
    $bed_number = $_POST["bed_number"]; 
    $notes = $_POST["notes"]; 
    $time = $_POST["time"]; 

    $mysql_qry = "insert into patients2 
    (patient_name, doctor_name, check_in_date, room_number, bed_number, notes, time) 
    values ('$patient_name', '$doctor_name', '$check_in_date', '$room_number', '$bed_number', '$notes', '$time';);     

    if($conn->query($mysql_qry) === TRUE) { 
     echo "Insert successful"; 
    } 
    else{ 
     echo "Error: " . $mysql_qry . "<br>" . $conn->error; 
    } 
    $conn->close(); 
?> 

Answer 1

你错过了封闭双引号，并有一个额外的分号：

这样的：

$mysql_qry = "insert into patients2 
    (patient_name, doctor_name, check_in_date, room_number, bed_number, notes, time) 
    values ('$patient_name', '$doctor_name', '$check_in_date', '$room_number', '$bed_number', '$notes', '$time';); 

需要如下：

$mysql_qry = "insert into patients2 
    (patient_name, doctor_name, check_in_date, room_number, bed_number, notes, time) 
    values ('$patient_name', '$doctor_name', '$check_in_date', '$room_number', '$bed_number', '$notes', '$time')"; 
    //                           ^

Answer 2

您使用额外的;在你的查询中。它不应该在那里。 您的查询应该是这样的：

$mysql_qry = "insert into patients2 
(patient_name, doctor_name, check_in_date, room_number, bed_number, notes, time) 
values ('$patient_name', '$doctor_name', '$check_in_date', '$room_number', '$bed_number', '$notes', '$time')";