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数据未能修改,感觉SQL出了问题。

qunxingw 发布于 2018-04-30 00:14, 2243 次点击


   
程序代码:
$conn = mysqli_connect($servername, $username, $password, $dbname);
mysqli_set_charset($conn, "utf8");
// Check connection
    if (!$conn) {
    die("连接失败: " . mysqli_connect_error());
    }

 

 

 if(isset($_POST['inoutid'])){
    $inoutid=$_POST['inoutid'];
    echo "所要修改数据的ID是:".$inoutid;
    $dept=$_POST['dept'];
    $chemnameid = $_POST['chemnameid'];
    $numbers =$_POST['numbers'];
    $date=$_POST['date'];
    $remarks= $_POST['remarks'];

 

 $sqlupdate="UPDATE table_cheminout SET chemnameid=$chemnameid,numbers=$numbers,dept=$dept,date=$date,remarks=$remarks WHERE inoutid=$inoutid";

 mysqli_query($conn ,$sqlupdate);
echo "<br>修改成功";

 }
mysqli_close($conn);

?>

<div id=form>

<div id=formin>
<h>修改此ID的数据</h>

<form action="<?php echo $_SERVER['PHP_SELF']; ?>" method="POST">
待修改ID<input type="number" name="inoutid"><br>

 部&nbsp&nbsp&nbsp&nbsp门:<select name="dept">
        <option value="ZY">ZY</option>
        <option value="YY">YY</option>
        </select ><br>

 化学品ID<input type="number" name="chemnameid"><br>

 数&nbsp&nbsp&nbsp&nbsp量:<input type="number" name="numbers"><br>

 日&nbsp&nbsp&nbsp&nbsp期: <input type="date" name="date"><br>

 备&nbsp&nbsp&nbsp&nbsp注:<input type="text" name="remarks" ><br>
  <input type="reset" value="重置">
  <input type="submit" value="提交">
</form>

1 回复
#2
qunxingw2018-05-01 08:08
$sqlupdate="UPDATE table_cheminout SET chemnameid=$chemnameid,numbers=$numbers,dept='$dept',date='$date',remarks='$remarks'
 WHERE inoutid=$inoutid";
1