不理解 mysqli_fetch_array

Question

所以我想我不明白这一点，我通常是编码的新手。我已经搜索并找不到一个很好的解释来让它在我的情况下工作。我需要从 fetch 命令中填充一个表，然后将每个结果与输入的信息更新到一个新列中。这是我的代码：

这填满了表格：

    echo "<table border='1'>
    <tr>
    <th>Envelope</th>
    <th>Budget</th>
    <th>Amount</th>
    </tr>";

    while($row = mysqli_fetch_array($result)) {

    echo "<tr>";
    echo "<td>" . $row['envelopename'] . "</td>";
    echo "<td>" . $row['envelopebudget'] . "</td>";
    ?><td><input type="text" name="budgetamount"></td><?php;
    echo "</tr>";
    }

    echo "</table>";
?>  
    <input type="submit">

这会将budgetname 的输入写入budgetname 列：

$paycheckname = mysqli_real_escape_string($con, $_POST['paycheckname']);
$budgetamount = mysqli_real_escape_string($con, $_POST['budgetamount']);
$envelopename = mysqli_real_escape_string($con, $_POST['envelopename']);
}

$sql="UPDATE envelopes SET $paycheckname='$budgetamount' WHERE envelopename ='$envelopename'";

if (!mysqli_query($con,$sql)) {
  die('Error: ' . mysqli_error($con));
}

现在它只是不写任何东西。我使用 echo var_dump($envelopename)."<br>"; and echo var_dump($budgetamount)."<br>"; 尝试查看它在做什么，但 $envelopename 始终为空白。感谢您的任何帮助，您可以提供。

Answer 1

只有<input>、<textarea>、<select>和<button>以表单形式提交给服务器

如果您想重新提交一些静态值，请创建一些隐藏输入

while($row = mysqli_fetch_array($result)) {

echo "<input type='hidden' value='$row[envelopename]' name='envelopename'/>";
echo "<input type='hidden' value='$row[envelopebudget]' name='envelopebudget'/>";
echo "<tr>";
echo "<td>" . $row['envelopename'] . "</td>";
echo "<td>" . $row['envelopebudget'] . "</td>";
?><td><input type="text" name="budgetamount"></td><?php;
echo "</tr>";
}

echo "</table>";

但您的下一个问题是您在循环中创建元素，因此您将有多个具有相同名称的输入

<tr>
    <input type='hidden' value='envName1' name='envelopename'/>
    <input type='hidden' value='envBudget1' name='envelopename'/>
    <td>envName1</td>
    <td>envBudget1</td>
    <td><input type="text" name="budgetamount"></td>
</tr>
<tr>
    <input type='hidden' value='envName2' name='envelopename'/>
    <input type='hidden' value='envBudget2' name='envelopename'/>
    <td>envName2</td>
    <td>envBudget2</td>
    <td><input type="text" name="budgetamount"></td>
</tr>

所以你必须以数组的形式提交

while($row = mysqli_fetch_array($result)) {

echo "<input type='hidden' value='$row[envelopename]' name='envelopename[]'/>";
echo "<input type='hidden' value='$row[envelopebudget]' name='envelopebudget[]'/>";
echo "<tr>";
echo "<td>" . $row['envelopename'] . "</td>";
echo "<td>" . $row['envelopebudget'] . "</td>";
?><td><input type="text" name="budgetamount[]"></td><?php;
echo "</tr>";
}

echo "</table>";

并在服务器端进程作为数组

 foreach ($_POST['budgetamount'] as $budgetamount){
       echo $budgetamount. '<br>';
  }

Answer 2

假设呈现的表单如下所示：

<tr>
    <input type='hidden' value='envName1' name='envelopename[]'/>
    <input type='hidden' value='envBudget1' name='envelopebudget[]'/>
    <td>envName1</td>
    <td>envBudget1</td>
    <td><input type="text" name="budgetamount[]"></td>
</tr>
<tr>
    <input type='hidden' value='envName2' name='envelopename[]'/>
    <input type='hidden' value='envBudget2' name='envelopebudget[]'/>
    <td>envName2</td>
    <td>envBudget2</td>
    <td><input type="text" name="budgetamount[]"></td>
</tr>
<tr>
    <input type='hidden' value='envName3' name='envelopename[]'/>
    <input type='hidden' value='envBudget3' name='envelopebudget[]'/>
    <td>envName3</td>
    <td>envBudget3</td>
    <td><input type="text" name="budgetamount[]"></td>
</tr>

当用户点击提交时，到达newpaycheck.php 的$_POST 将如下所示：

$_POST
['envelopename']{
   [0]=>'envName1',
   [1]=>'envName2',
   [2]=>'envName3'
},['envelopebudget']{
   [0]=>'envBudget1',
   [1]=>'envBudget2',
   [2]=>'envBudget3'
},['budgetamount']{
   [0]=>'someValueEnteredByUser',
   [1]=>'anotherValueEnteredByUser',
   [2]=>'yetAnotherValueEnteredByUser'
}

所以你可以这样做：

foreach ($_POST['envelopename'] as $envelopename){
   $arrayIndex = array_search($envelopename,$_POST['envelopename']);
   $envelopebudget = $_POST['envelopebudget'][$arrayIndex];
   $budgetamount= $_POST['budgetamount'][$arrayIndex];

   $paycheckname = mysqli_real_escape_string($con, $envelopebudget);
   $budgetamount = mysqli_real_escape_string($con,$budgetamount);
   $envelopename = mysqli_real_escape_string($con,$envelopename);

   $sql="UPDATE envelopes SET $paycheckname='$budgetamount' WHERE envelopename ='$envelopename'";

   if (!mysqli_query($con,$sql)) {
       die('Error: ' . mysqli_error($con));
    }

}