我的php表单只保存表的最后一行

hs1ihplo 于 2021-06-25 发布在 Mysql

关注(0)|答案(3)|浏览(209)

我在表单中有一个表（new\u tear.php），它从单个用户获取数据。第一个cloumn是从具有名称的数据库表（profiletable）填充的。用户必须从下拉列表中为每一行（每个名称）选择一个选项。提交表单（post to tear\u done.php）时，我希望将每一行存储在数据库中。但是，我每次只看到最后一行
新建\u ter.php

<?php
$sqlT1 = "SELECT * FROM ProfileTable WHERE dept = 'T1'";
$queryT1 = mysqli_query($conn, $sqlT1);
if (!$queryT1) {
  die ('SQL Error: ' . mysqli_error($conn));
}
?>
<form action="tear_done.php" method="post">
<table>
<?php
    $no     = 1;
    while ($row = mysqli_fetch_array($queryT1))
    {
        echo '<tr>
                <td>'.$no.'</td>
                <td>'.$row['name'].'</td>
                <td><select id="Dz_M" name="Dz_M">
                    <option value="4pm">4pm</option>
                    <option value="8am">8am</option>
                    <option value="Rest">Rest</option>
                </select><br>
                </td>
                <td><select id="Dz_T" name="Dz_T">
                    <option value="4pm">4pm</option>
                    <option value="8am">8am</option>
                    <option value="Rest">Rest</option>
                </select><br>
                </td>
                <td><select id="Dz_W" name="Dz_W">
                    <option value="4pm">4pm</option>
                    <option value="8am">8am</option>
                    <option value="Rest">Rest</option>
                </select><br>
                </td>
                <td><select id="Dz_F" name="Dz_F">
                    <option value="4pm">4pm</option>
                    <option value="8am">8am</option>
                    <option value="Rest">Rest</option>
                </select><br>
                </td>
                <td><select id="Dz_S" name="Dz_S">
                    <option value="4pm">4pm</option>
                    <option value="8am">8am</option>
                    <option value="Rest">Rest</option>
                </select><br>
                </td>                   
                </tr>';
        $no++;
        $no_cont = $no;
    }?> 
</table>
<br>
<button value="save" type="submit" name="save">Save Tear</button>       
</form>

撕裂完成.php

<?php
$sqlT1 = "SELECT * FROM ProfileTable WHERE dept = 'T1'";
$queryT1 = mysqli_query($conn, $sqlT1);
if (!$queryT1) {
    die ('SQL Error: ' . mysqli_error($conn));
}
$no = 1;
while ($row = mysqli_fetch_array($queryT1))
{
$Name = $row ['name'];
$Dept = $row ['dept'];
$DZ_M = $_POST ["Dz_M"];
$DZ_T = $_POST ["Dz_T"];
$DZ_W = $_POST ["Dz_W"];
$DZ_F = $_POST ["Dz_F"];
$DZ_S = $_POST ["Dz_S"];

$sql = "INSERT INTO TearTable (name, Mdy, Tdy, Wdy, Fdy, Sdy, dept)
VALUES ('$Name', '$DZ_M', '$DZ_T', '$DZ_W', '$DZ_F', '$DZ_S', '$Dept')";    

echo $no;
echo '<br>';
echo $sql; //for testing
echo '<br>';
$no++;
}   
?>

我从echo得到的是所有名称和dept都正确的sql，但是其他字段$dzïm，$dzït等只重复从newïtear.php表中输入的最后一行。请问我遗漏了什么？

mysql php Html

来源：https://stackoverflow.com/questions/50278477/my-php-form-is-only-saving-the-last-row-of-my-table

3条答案

按热度按时间

lnxxn5zx1#

最好是创建一个大查询语句。

$sql .= "INSERT INTO TearTable (name, Mdy, Tdy, Wdy, Fdy, Sdy, dept) VALUES"
while ($row = mysqli_fetch_array($queryT1)) {  
  $Name = $row ['name'];
  $Dept = $row ['dept'];
  $DZ_M = $_POST ["Dz_M"];
  $DZ_T = $_POST ["Dz_T"];
  $DZ_W = $_POST ["Dz_W"];
  $DZ_F = $_POST ["Dz_F"];
  $DZ_S = $_POST ["Dz_S"]; 
  $sql .= "('$Name', '$DZ_M', '$DZ_T', '$DZ_W', '$DZ_F', '$DZ_S', '$Dept'),"
}
$result = mysqli_query($conn, $sql); //run your INSERT query.

这个查询将所有数据插入数据库。这种方法的问题是，如果insert失败，调试就会变得更加困难。

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r7knjye22#

我可以通过在新的\u tear.php select id和name中添加var（$no）来解决我的问题。
新建\u tear.php

<?php
    $no     = 1;
    while ($row = mysqli_fetch_array($queryT1))
    {
        echo '<tr>
                <td>'.$no.'</td>
                <td>'.$row['name'].'</td>
                <td><select id="Dz_M'  . $no . '" name="Dz_M'  . $no . '">
                    <option value="4pm">4pm</option>
                    <option value="8am">8am</option>
                    <option value="Rest">Rest</option>
                </select><br>
                </td>
                <td><select id="Dz_T'  . $no . '" name="Dz_T'  . $no . '">
                    <option value="4pm">4pm</option>
                    <option value="8am">8am</option>
                    <option value="Rest">Rest</option>
                </select><br>
                </td>
                <td><select id="Dz_W'  . $no . '" name="Dz_W'  . $no . '">
                    <option value="4pm">4pm</option>
                    <option value="8am">8am</option>
                    <option value="Rest">Rest</option>
                </select><br>
                </td>
                <td><select id="Dz_F'  . $no . '" name="Dz_F'  . $no . '">
                    <option value="4pm">4pm</option>
                    <option value="8am">8am</option>
                    <option value="Rest">Rest</option>
                </select><br>

                </td>
                <td><select id="Dz_S'  . $no . '" name="Dz_S'  . $no . '">
                    <option value="4pm">4pm</option>
                    <option value="8am">8am</option>
                    <option value="Rest">Rest</option>
                </select><br>
                </td>                   
                </tr>';
        $no++;
        $no_cont = $no;
    }?>

并将var（$no）添加到tear\u done.php中的$\u post中
撕裂完成.php

$Name = $row ['name'];
$Dept = $row ['dept'];
$DZ_M = $_POST ['Dz_M'.$no]; 
$DZ_T = $_POST ['Dz_T'.$no];
$DZ_W = $_POST ['Dz_W'.$no];
$DZ_F = $_POST ['Dz_F'.$no];
$DZ_S = $_POST ['Dz_S'.$no];

这解决了我的问题，现在在echo中（同样，仅用于测试），我看到sql看起来是正确的。谢谢大家的帮助。

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46scxncf3#

在while循环中插入查询字符串

while ($row = mysqli_fetch_array($queryT1))
{
......

$sql .= "INSERT INTO TearTable (name, Mdy, Tdy, Wdy, Fdy, Sdy, dept)
VALUES ('$Name', '$DZ_M', '$DZ_T', '$DZ_W', '$DZ_F', '$DZ_S', '$Dept');";    // Note .= and semicolon at the end 

   .....
    } 
// Outside the loop
execute your query here..

出于安全考虑，你应该使用事先准备好的声明

赞(0）回复(0）举报 2021-06-25

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我的php表单只保存表的最后一行

3条答案

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