Question

我是PHP的新手，正在编写一本地址簿，其中包含所述公司的公司和联系人。当点击提交以更新数据时，它只是用最初在那里的旧数据替换它，这里是公司的html表单：

<html>
<head><title>Update Records in MYSQL Database</title>
    <style>
        body {background-color: powderblue;}
        form {border: 5px solid midnightblue;
            padding: 40px;}
        label {color: darkblue;}
        form {border: 5px solid midnightblue;
            padding: 40px;
    </style>
</head>
<body>
<?php
    //Connect to the Database
    $link = mysqli_connect('localhost', '', '');
    //Select the Database
    mysqli_select_db($link, 'ADDRESS_BOOK_DB');
    //Select Query
    $sql = "SELECT * FROM Companies";
    //Execute the Query
    $records = mysqli_query($link, $sql);
?>
<form>
    <table>
        <tr>
            <th>CompanyName</th>
            <th>Address</th>
            <th>City</th>
            <th>State</th>
            <th>ZipCode</th>
            <th>PhoneNumber</th>
        </tr>
        <?php
        while($row = mysqli_fetch_array($records)) {
            echo "<tr><form action=update.php method='post'></form>";
            echo "<td><input type='text' name='CompanyName' value='" . $row['companyname1'] . "'></td>";
            echo "<td><input type='text' name='address' value='" . $row['address'] . "'></td>";
            echo "<td><input type='text' name='city' value='" . $row['city'] . "'></td>";
            echo "<td><input type='text' name='state' value='" . $row['state'] . "'></td>";
            echo "<td><input type='text' name='zipcode' value='" . $row['zipcode'] . "'></td>";
            echo "<td><input type='text' name='phone' value='" . $row['phone'] . "'></td>";
            echo "<td><input type=hidden name=companies_id value='" . $row['companies_id'] . "'>";
            echo "<td><input type='submit'>";
            echo "</form></tr>";
        }
    ?>
    </table>
</form>
</body>
</html>

这里是实际更新的php sql查询无法正常工作或任何输入有助于摆脱我的头痛。

<?php
//Connect to the Database
$link = mysqli_connect('localhost', '', '');
//Select the Database
mysqli_select_db($link, 'ADDRESS_BOOK_DB');

//Update Query
$sql = "UPDATE Companies SET CompanyName= $_POST[companyname1], Address= $_POST[address], City= $_POST[city], State= $_POST[state], ZipCode= $_POST[zipcode], PhoneNumber= $_POST[phone] 
WHERE companies_id= $_POST[companies_id]";

//Execute the Query
if(mysqli_query($link, $sql))
    header("refresh:1; url=edit.php");
    enter code here

else
    echo "Not Updated";
?>

Answer 1

首先将此条件与WHERE子句$sql = "SELECT * FROM Companies";放在一起，您希望更新哪个数据库记录。看起来你正在从form内的数据库中获取所有记录

其次你在while循环的begening处关闭了你的form标记并保留了其余的输入字段没有form

改变这个

while($row = mysqli_fetch_array($records)) {
        echo "<tr><form action=update.php method='post'></form>";
        echo "<td><input type='text' name='CompanyName' value='" . $row['companyname1'] . "'></td>";
        echo "<td><input type='text' name='address' value='" . $row['address'] . "'></td>";
        echo "<td><input type='text' name='city' value='" . $row['city'] . "'></td>";
        echo "<td><input type='text' name='state' value='" . $row['state'] . "'></td>";
        echo "<td><input type='text' name='zipcode' value='" . $row['zipcode'] . "'></td>";
        echo "<td><input type='text' name='phone' value='" . $row['phone'] . "'></td>";
        echo "<td><input type=hidden name=companies_id value='" . $row['companies_id'] . "'>";
        echo "<td><input type='submit'>";
        echo "</form></tr>";
    }

到

?>
<form action=update.php method='post'>
<?php
while($row = mysqli_fetch_array($records)) {

        echo "<td><input type='text' name='CompanyName' value='" . $row['companyname1'] . "'></td>";
        echo "<td><input type='text' name='address' value='" . $row['address'] . "'></td>";
        echo "<td><input type='text' name='city' value='" . $row['city'] . "'></td>";
        echo "<td><input type='text' name='state' value='" . $row['state'] . "'></td>";
        echo "<td><input type='text' name='zipcode' value='" . $row['zipcode'] . "'></td>";
        echo "<td><input type='text' name='phone' value='" . $row['phone'] . "'></td>";
        echo "<td><input type=hidden name=companies_id value='" . $row['companies_id'] . "'>";
        echo "<td><input type='submit'>";

    }

?>
</form>

使用php问题的mysql更新表单不断更新旧信息而不是新的更新

1 个答案: