Php 获取按钮的id值

Php 获取按钮的id值,php,mysql,database,Php,Mysql,Database,这是我代码的一部分。我想在单击submit时获取按钮的id作为PersonID的值。这样我就可以根据PersonID从数据库中获取单独的id值 while($row = mysql_fetch_array($result, MYSQL_ASSOC)) { print "<tr> <td>"; echo $row['PersonID']; print "</td> <td>"; echo $row['FirstNam

这是我代码的一部分。我想在单击submit时获取按钮的id作为PersonID的值。这样我就可以根据PersonID从数据库中获取单独的id值

while($row = mysql_fetch_array($result, MYSQL_ASSOC))
{   
print "<tr> <td>";

    echo $row['PersonID']; 
    print "</td> <td>";
    echo $row['FirstName']; 
    print "</td> <td>";
    echo $row['LastName']; 
    print "</td> <td>";
    echo $row['Address'];
    print "</td> <td>";
    echo "<input id=\"<?php echo $row['PersonID'];?>\" type=\"submit\" name=\"btn-signup\" value=\"Submit\">\n";

    print "</td> </tr>";


}
while($row=mysql\u fetch\u数组($result,mysql\u ASSOC))
{   
打印“”;
echo$row['PersonID'];
打印“”;
echo$row['FirstName'];
打印“”;
echo$row['LastName'];
打印“”;
echo$row['Address'];
打印“”;

echo“您可以在带引号的文本(双引号)中使用大括号来访问某些类型的变量,而不需要复杂的转义,这使得通过转义引号更容易编写和发现错误

while( $row = mysql_fetch_array( $result, MYSQL_ASSOC ) ){   
    echo "
    <tr>
        <td>{$row['PersonID']}</td>
        <td>{$row['FirstName']}</td>
        <td>{$row['LastName']}</td>
        <td>{$row['Address']}</td>
        <td>
            <input type='submit' name='btn-signup' value='Submit'>
            <input type='hidden' name='personID' value='{$row['PersonID']}' />
        </td>
    </tr>";
}
while($row=mysql\u fetch\u数组($result,mysql\u ASSOC)){
回声“
{$row['PersonID']}
{$row['FirstName']}
{$row['LastName']}
{$row['Address']}
";
}

我想您希望在单击提交按钮时获取id

您应该将其设置为隐藏字段,并将id指定为该字段的值,而不是采用这种方法

另外,您应该遵循干净的编码模式,避免在echo语句中编写HTML代码。将其分解为单独的HTML和PHP标记

<?php while($row = mysql_fetch_array($result, MYSQL_ASSOC)) { ?> 
<tr> 
   <td> <?php echo $row['PersonID']; ?> </td> 
   <td> <?php echo $row['FirstName']; ?> </td> 
   <td> <?php echo $row['LastName']; ?>  </td> 
   <td> <?php echo $row['Address']; ?>   </td> 
   <td>
     <input type="hidden" name="id" value="<?php echo $row['PersonID'];?>"> <!-- Hidden field containing ID -->
     <input type="submit" name="submit" value="Submit">
   </td> 
</tr>
<?php } ?>

请在下面试试

注意:确保在代码中定义了
标记

while($row = mysql_fetch_array($result, MYSQL_ASSOC)) {   
    print "<tr> <td>";
    echo $row['PersonID']; 
    print "</td> <td>";
    echo $row['FirstName']; 
    print "</td> <td>";
    echo $row['LastName']; 
    print "</td> <td>";
    echo $row['Address'];
    print "</td> <td>";
    echo '<input type="submit" name="btn-signup" value="Submit">';
    echo '<input type="hidden" name="PersonID" value="'.$row['PersonID'].'">';
    print "</td> </tr>";
}
while($row=mysql\u fetch\u数组($result,mysql\u ASSOC)){
打印“”;
echo$row['PersonID'];
打印“”;
echo$row['FirstName'];
打印“”;
echo$row['LastName'];
打印“”;
echo$row['Address'];
打印“”;
回声';
回声';
打印“”;
}

与将id传递给submit按钮的valueecho不同,“基本上,您可以使用字符串连接和.echo之类的运算符,我将获得作为id的值,但我希望按钮的行为与我单击submit时相同,它应该获得整个数据库表的值id@stefen我这样做了,但它显示出错误