Question

我的PHP表单有问题，应该更新我的MySQL数据库表的一个人的年龄。该列名为Age。我得到的问题是当我尝试更新它不会更新并显示的年龄时

UPDATE Persons SET Age=Array WHERE FirstName ='Array'
array(13) { [0]=> string(0) "" [1]=> string(0) "" [2]=> string(0) "" [3]=> string(0) "" [4]=> string(0) "" [5]=> string(0) "" [6]=> string(0) "" [7]=> string(0) "" [8]=> string(0) "" [9]=> string(0) "" [10]=> string(0) "" [11]=> string(0) "" [12]=> string(3) "544" } Cannot update

在我的页面底部。您可以在http://thetotempole.ca/phptester/editpage.php

查看我的页面

我页面的所有代码：

<?php
    $hostname = "";//host name
    $dbname = "";//database name
    $username = "";//username you use to login to php my admin
    $password = "";//password you use to login

    //CONNECTION OBJECT
    //This Keeps the Connection to the Databade
    $conn = new MySQLi($hostname, $username, $password, $dbname) or die('Can not connect to database')      
?>
<!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Transitional//EN" "http://www.w3.org/TR/xhtml1/DTD/xhtml1-transitional.dtd">
<html xmlns="http://www.w3.org/1999/xhtml">
<head>
<meta http-equiv="Content-Type" content="text/html; charset=UTF-8" />
<title>Untitled Document</title>
</head>
<body>
<?php

//$id=$_GET['FirstName'];

//Create a query
$sql = "SELECT * FROM Persons";
//submit the query and capture the result
$result = $conn->query($sql) or die(mysql_error());
$query=getenv(QUERY_STRING);
parse_str($query);
?>
<h2>Update Record <?php echo $sql;?></h2>
<a href="cool.php">Click Here to see the Table Data</a>
<?php $counter = 0; ?>
<form action="" method="post">
<?php

while ($row = $result->fetch_assoc()) {?>

<table border="0" cellspacing="10">
<tr>
<td>Age:</td>
<td><?php echo $row['Age'][$counter];?></td> 
<td><?php echo $row['FirstName'];?></td>
<td><input type="text" name="Age[]"></td>
<td><input type="text" name="FirstName[]" value="<?php echo $row['FirstName'];?>" /></td>
</tr>

<tr>
<td><INPUT TYPE="Submit" VALUE="Update the Record" NAME="Submit"></td>
</tr>
</table>
<?php
$counter++;   }
?>
</form>

<?php
if(isset($_POST['Submit'])){//if the submit button is clicked
$age = $_POST['Age'];
$id = $_POST['FirstName'];

$sql="UPDATE Persons SET Age=$age WHERE FirstName ='$id'";
echo $sql."<br>";
var_dump($_POST[Age]);
$conn->query($sql) or die("Cannot update");//update or error
}
?>

</body>
</html>

我将代码更新为我当前的代码。

Answer 1

您可以使用数组：

<td><input type="text" name="person[<?php echo $row['FirstName'];?>]"></td>

和

if(isset($_POST['Submit']))
foreach ($_POST['person'] as $id => $age)
{
  $sql="UPDATE Persons SET Age='".$age."' WHERE firstname ='".$id."'";
  $conn->query($sql) or die("Cannot update");//update or error
}

另外，不要忘记表单输入名称中FirstName中的mysql字符串转义和html实体。

Answer 2

以下是phpfiddle的内容，以便于访问：

<?php
if(isset($_POST['Submit'])){ //if ANY submit button has been clicked
    if (is_numeric($_POST['Age'])) {
        $age = (int) $_POST['Age'];
    } elseif ($age == '') {
        $age = (string) 'NULL';
        die('Please go back and enter a valid age.');
    } else { 
        die('Please go back and enter a valid age.');
    }
    $id = mysqli_real_escape_string($conn,$_POST['FirstName']);
    $sql="UPDATE Persons SET Age=$age WHERE FirstName ='$id'";
    $conn->query($sql) or die("Cannot update"); //update or error
}
$sql="SELECT * FROM Persons";
$conn->query($sql) or die("Cannot update"); //update or error
while ($row = $result->fetch_assoc()){
    $age = (int) $row['Age'];
    $id = mysqli_real_escape_string($conn,$row['FirstName']);
?>
    <form action="" method="post">
        <table border="0" cellspacing="10">
            <tr>
                <td>Age:</td>
                <td><?php echo $age;?></td> 
                <td><?php echo $id;?></td>
                <td><input type="text" name="Age"></td>
                <td><input type="text" name="FirstName" value="<?php echo $id;?>"><?php echo $id;?></input></td>
            </tr>
            <tr>
                <td><INPUT TYPE="Submit" VALUE="Update the Record" NAME="Submit"></td>
            </tr>
        </table>
    </form>
<?php
}
?>

如果您的变量是数字，那么更新语句不需要单引号。我只纠正了Age，因为我不确定名字。此外，PHP不需要将变量串联到由双引号括起来的字符串中。

希望这有帮助。

编辑：当您转义字符串时，请务必使用函数的mysqli版本而不是mysql，因为后者已被弃用。

PHP表单仅更新MySQL数据库表的底部输入

2 个答案: