通过提交显示所有行并更新所有行

Question

我想通过 php 查询显示所有行，并通过 sql 中的提交按钮更新所有行。我以这种方式在下面可以显示所有行并通过其自己的提交按钮更新特定行。但我想通过一个提交按钮来更新所有内容。

所以，谢谢，我想循环更新。但我不明白在这种情况下该怎么做。

这是我的代码：

<?php
include_once('../db.php');
global $db;
$result = mysqli_query($dbh,"SELECT * FROM ppad");
if(!$result) {
    die("Database query failed: " . mysqli_error());
}
    while($row = mysqli_fetch_assoc($result)) {
    $id=$row['id']; 
    $name=$row['name'];
    $date=$row['date'];
    $title=$row['title'];
    $Detail=$row['Detail'];

echo '<form action="padSproccess.php" method="POST">        
<table width="100%" border="1">
<tr>
    <td width="10%">Date</td>
    <td width="14%">Time</td>
    <td width="20%">Name(url)</td>
    <td width="30%">Detail</td>
</tr>
<tr>
    <td width="10%"><input type="text" name="date" maxlength="2" value="'.$date.'"></td>
    <td width="14%"><input type="text" name="title" maxlength="50" value="'.$title.'"></td>
    <td width="20%"><input type="text" name="name" maxlength="50" value="'.$name.'"></td>
    <td width="30%"><input type="text" name="Detail" maxlength="100" value="'.$Detail.'"></td>
    <input type="hidden" name="id" value="'.$id.'">
    </tr>
</table>
    <input type="submit" name="submit" id="submit" value="Submit">
</form>';}
?>

padSproccess.php

include("../db.php");
global $db;
if(isset($_POST['submit'])){
    $date = mysqli_real_escape_string($dbh,$_POST['date']);
    $title = mysqli_real_escape_string($dbh,$_POST['title']);
    $name = mysqli_real_escape_string($dbh,$_POST['name']);
    $Detail = mysqli_real_escape_string($dbh,$_POST['Detail']);
    $id = mysqli_real_escape_string($dbh,$_POST['id']);
    // update data in mysql database 
    $update = mysqli_query($dbh,"UPDATE ppad SET date='$date', month='$month', name='$name', Detail='$Detail' WHERE id = '$id'"); 

// if successfully updated. 
}

Answer 1

您首先需要做的是有一个整体的表单，而不是每个表单（除非您想使用 javascript 来触发 ajax 调用）。所以你需要做的是确保每一行都可以与一个特定的 id 相关联：

<?php

include_once '../db.php';

$result = mysqli_query($dbh, "SELECT * FROM ppad");

if(!$result) {
    die("Database query failed: " . mysqli_error());
}

?>

<form action="padSproccess.php" method="POST">
    <table width="100%" border="1">
        <thead>
            <tr>
                <td width="10%">Date</td>
                <td width="14%">Time</td>
                <td width="20%">Name(url)</td>
                <td width="30%">Detail</td>
            </tr>
        </thead>
        <tbody>
        <?php

            while($row = mysqli_fetch_assoc($result)) {
                $id     = $row['id']; 
                $name   = $row['name'];
                $date   = $row['date'];
                $title  = $row['title'];
                $Detail = $row['Detail'];

                echo '
                    <tr>
                        <td width="10%"><input type="text" name="date[' . $id . ']" maxlength="2" value="'.$date.'"></td>
                        <td width="14%"><input type="text" name="title[' . $id . ']" maxlength="50" value="'.$title.'"></td>
                        <td width="20%"><input type="text" name="name[' . $id . ']" maxlength="50" value="'.$name.'"></td>
                        <td width="30%"><input type="text" name="Detail[' . $id . ']" maxlength="100" value="'.$Detail.'"></td>
                    </tr>
                ';
            }

        ?>
        </tbody>
    </table>
    <input type="submit" name="submit" id="submit" value="Submit">
</form>

然后在padSproccess.php 中，您将收到names、dates、titles 和Details 的数组，每一个都由行的ID 键入。所以会变成这样：

<?php

include "../db.php";

if(isset($_POST['submit'])){
    $ids = array_keys($_POST['name']);

    foreach ($ids as $id) {
        $date   = mysqli_real_escape_string($dbh,$_POST['date'][$id]);
        $title  = mysqli_real_escape_string($dbh,$_POST['title'][$id]);
        $name   = mysqli_real_escape_string($dbh,$_POST['name'][$id]);
        $Detail = mysqli_real_escape_string($dbh,$_POST['Detail'][$id]);
        $id     = mysqli_real_escape_string($id);
        // update data in mysql database 
        $update = mysqli_query($dbh,"UPDATE ppad SET date='$date', month='$month', name='$name', Detail='$Detail' WHERE id = '$id'"); 
    }

// if successfully updated. 
}

Answer 2

试试这个。当然对于我还没有启动的数据库，可能会出错。

<?php
include_once('../db.php');
global $db;
$result = mysqli_query($dbh,"SELECT * FROM ppad");
if(!$result) {
    die("Database query failed: " . mysqli_error());
}
?>
<form action="padSproccess.php" method="POST">
<?php
    while($row = mysqli_fetch_assoc($result)) {
    $id=$row['id']; 
    $name=$row['name'];
    $date=$row['date'];
    $title=$row['title'];
    $Detail=$row['Detail'];

echo '        
<table width="100%" border="1">
<tr>
    <td width="10%">Date</td>
    <td width="14%">Time</td>
    <td width="20%">Name(url)</td>
    <td width="30%">Detail</td>
</tr>
<tr>
    <td width="10%"><input type="text" name="ar['.$id.'][date]" maxlength="2" value="'.$date.'"></td>
    <td width="14%"><input type="text" name="ar['.$id.'][title]" maxlength="50" value="'.$title.'"></td>
    <td width="20%"><input type="text" name="ar['.$id.'][name]" maxlength="50" value="'.$name.'"></td>
    <td width="30%"><input type="text" name="ar['.$id.'][Detail]" maxlength="100" value="'.$Detail.'"></td>
    </tr>
</table>

';}
    ?>
    <input type="submit" name="submit" id="submit" value="Submit">
    </form>

<?php   
include("../db.php");
global $db;
if(isset($_POST['submit'])){
    foreach($_POST['ar'] as $id=>$dat){
        $date = mysqli_real_escape_string($dbh,$dat['date']);
        $title = mysqli_real_escape_string($dbh,$dat['title']);
        $name = mysqli_real_escape_string($dbh,$dat['name']);
        $Detail = mysqli_real_escape_string($dbh,$dat['Detail']);
        $id = mysqli_real_escape_string($dbh,$id]);
        // update data in mysql database 
        $update = mysqli_query($dbh,"UPDATE ppad SET date='$date', month='$month', name='$name', Detail='$Detail' WHERE id = '$id'"); 
    }
// if successfully updated. 
}
?>

Answer 3

为此，您需要将代码更新为

<?php
include_once('../db.php');
global $db;
$result = mysqli_query($dbh,"SELECT * FROM ppad");
if(!$result) {
    die("Database query failed: " . mysqli_error());
}?>
<form action="padSproccess.php" method="POST">        
<table width="100%" border="1">
<tr>
    <td width="10%">Date</td>
    <td width="14%">Time</td>
    <td width="20%">Name(url)</td>
    <td width="30%">Detail</td>
</tr>
<?php 
    while($row = mysqli_fetch_assoc($result)) {
    $id=$row['id']; 
    $name=$row['name'];
    $date=$row['date'];
    $title=$row['title'];
    $Detail=$row['Detail'];

echo '<tr>
    <td width="10%"><input type="text" name="date[]" maxlength="2" value="'.$date.'"></td>
    <td width="14%"><input type="text" name="title[]" maxlength="50" value="'.$title.'"></td>
    <td width="20%"><input type="text" name="name[]" maxlength="50" value="'.$name.'"></td>
    <td width="30%"><input type="text" name="Detail[]" maxlength="100" value="'.$Detail.'"></td>
    <input type="hidden" name="id[]" value="'.$id.'">
    </tr>';
}?>

</table>
    <input type="submit" name="submit" id="submit" value="Submit">
</form>

现在，在您的 padSproccess.php 中，您将在变量中收到一组结果，这些结果将使用 foreach 循环进行更新