在查询中使用PHP变量时，MySQL错误1064

<?php
    session_start();
    $con = mysqli_connect("localhost","root","12369","medical");
    $data1 = $_SESSION["symp1"];
    $data2 = $_SESSION["symp2"];
    $data3 = $_SESSION["symp3"];
    $data4 = $_SESSION["symp4"];
    $finalData = implode(' ', array($data1, $data2, $data3, $data4));
    $userinput = $_REQUEST["answer"];
    $dname=$_SESSION["dname"];
    $dname = str_replace(' ', '_', $dname);
    echo $dname."  <br>";
    $sql = " UPDATE diseases SET UserInput = $finalData WHERE Name =   $dname ";
    if($userinput=='yes'){  
        if(mysqli_query($con,$sql)){
            echo "Values inserted";
            $_SESSION["info"] = "yes";
            header('Location: http://localhost/medical/last.php');
    }else{
            echo mysqli_errno($con);
            $_SESSION["info"] = "no";
            //header('Location: http://localhost/medical/last.php');
    }
   }
?>

I'm getting error 1064? I already read answers to similar question, but my code doesn't work. My table schema is:

我得到错误1064 ?我已经读过类似问题的答案，但是我的代码不起作用。我的表模式是:

CREATE TABLE IF NOT EXISTS `diseases` (
  `ID` int(50) NOT NULL AUTO_INCREMENT,
  `Name` varchar(255) NOT NULL,
  `Symptoms` varchar(255) NOT NULL,
  `Medicines` varchar(255) NOT NULL,
  `Description` varchar(255) NOT NULL,
  `Tags` varchar(255) NOT NULL,
  `UserInput` varchar(255) NOT NULL,
  PRIMARY KEY (`ID`)
)

What's wrong in my code? Thanks

我的代码有什么问题?谢谢

3 个解决方案

#1

Change:

变化:

$sql = " UPDATE diseases SET UserInput = $finalData WHERE Name =   $dname ";

to:

$sql = "UPDATE `diseases` SET `UserInput` = '$finalData' WHERE `Name` = '$dname'";

Add single quotes around variables that contain a string. Add backticks around columns and table to prevent mysql reserved words error

在包含字符串的变量周围添加单引号。在列和表周围添加回勾以防止mysql保留字错误

It would be even better to use mysqli_prepare do the following:

最好使用mysqli_prepare执行以下操作:

$stmt = mysqli_prepare($con, "UPDATE `diseases` SET `UserInput` = ? WHERE `Name` = ?");
mysqli_stmt_bind_param($stmt, "ss", $finalData, $dname);
mysqli_stmt_execute($stmt);

#2

As the error message should state, you have an error in your SQL syntax:

由于错误消息应该说明，您的SQL语法中有一个错误:

MySQL Error 1064: You have an error in your SQL syntax

MySQL错误1064:在SQL语法中有一个错误。

Surround your data by single quotes and you are good to go. Furthermore, Name is a reserved keyword in MySQL. You can still use it in your query, though, but you should consider escaping table names with backticks:

用单引号括住你的数据，你就可以开始了。此外，Name是MySQL中的保留关键字。不过，您仍然可以在查询中使用它，但您应该考虑使用带反勾号的转义表名:

$sql = " UPDATE diseases SET `UserInput` = '$finalData' WHERE `Name` = '$dname' ";

#3

Add single qoutes around your data:

在你的数据周围增加单独的qoutes:

 $sql = " UPDATE diseases SET UserInput = '$finalData' WHERE Name =   '$dname' ";

or better use prepared statements

或者更好地使用准备好的语句

#1