从下拉列表中显示数据php

问题描述 投票:0回答:3

我正努力在项目的下拉列表中显示所选选项的详细信息。

我有一个从MYSQLi查询填充的下拉列表。我希望用户选择一个选项,并从数据库中提取相关的值并显示给用户。

动态填充的下拉列表用于人员的“FirstName”(表名),并且当用户从下拉列表中选择名称时,我希望显示该人的记录。

下面的代码用于动态填充下拉列表。用户单击该按钮并转到下一页,该页面应创建包含结果的表。没有错误,但也没有要求的结果。

下拉代码

<!DOCTYPE>
<html>
<head>

<title>Update Data</title>
</head>

<body>

<form name="form_update" method="post" action="update_test.php">

<?php
$con=mysqli_connect("localhost","root","","ismat_db");
//============== check connection
if(mysqli_errno($con))
{
echo "Can't Connect to mySQL:".mysqli_connect_error();
}



//This creates the drop down box

echo "<select name= 'FirstName'>";
echo '<option value="">'.'--- Please Select Person ---'.'</option>';
$query = mysqli_query($con,"SELECT FirstName FROM persons");
$query_display = mysqli_query($con,"SELECT * FROM persons");

while($row=mysqli_fetch_array($query))
{
echo "<option value='". $row['id']."'>".$row['FirstName']
 .'</option>';

}

echo '</select>';

?>&nbsp;<input type="submit" name="submit" value="Submit"/>
</form><br/><br/>
<a href="main.html"> Go back to Main Page </a>
</body>
</html>

查看代码

<title>Update Data</title>
</head>

<body>
<!--<table>
<tr>
<td align="center"> From Database </td>
</tr>
<tr>
<td>
    <table border="1">
        <tr>
            <td>First Name</td>
            <td>Last Name </td>
            <td> Gender </td>
            <td> Subject </td>
            <td> Hobbies </td>
        </tr>
     -->
  <?php 
$con=mysqli_connect("localhost","root","","ismat_db");
if(mysqli_errno($con))
{
echo "Can't Connect to mySQL:".mysqli_connect_error();
}


//$name = mysqli_real_escape_string($con,$_POST['select']);
 // $fetch = mysqli_query($con,"SELECT * FROM persons WHERE FirstName='".$name."'");
//  $row_display=mysqli_fetch_assoc($fetch);

if(isset($_POST['select']))
{

$name = mysqli_real_escape_string($con,$_POST['select']);
  $fetch = "SELECT * FROM persons WHERE FirstName = '".$name."'";
 $result = mysqli_query($con,$fetch);

 //display the table
 echo '<table border="1">'.'<tr>'.'<td align="center">'. 'From Database'. '</td>'.'</tr>';
echo '<tr>'.'<td>'.'<table border="1">'.'<tr>'.'<td>'.'First Name'.'</td>'
.'<td>'.'Last           Name'.'</td>'.'<td>'. 'Gender' .'</td>'.'<td>'
. 'Subject'. '</td>'.'<td>'. 'Hobbies' .'</td>'.'</tr>';
while($data = mysqli_fetch_row($result))
{
echo ("<tr><td>$data[0]</td><td>$data[1]</td><td>$data[2]</td><td>$data[3]</td><td>$data[4] </td></tr>");
 }

  echo '</table>'.'</td>'.'</tr>'.'</table>';
    }

   ?>
   <!--
   </table>
   </td>
   </tr>
   </table>
   -->
   <br/>
   <a href="update.php"> Go back to Main Page </a>
   </body>
   </html>
php mysqli
3个回答
2
投票

您在视图代码中使用了错误的fieldname:

echo "<select name= 'FirstName'>";
                     ^^^^^^^^^

V.S.

$name = mysqli_real_escape_string($con,$_POST['select']);
                                               ^^^^^^

表单是使用fieldname=value对提交的,而不是输入来自的标记的名称。


0
投票

建议很少

  1. 在您的查询中添加您设置的ID作为选项的值 echo "<option value='". $row['id']."'>".$row['FirstName'] 所以更新如下: $query = mysqli_query($con,"SELECT id,FirstName FROM persons");
  2. 您正在检查错误的POST var 'select',它应该是'FirstName',它是您在下面使用的选择字段的名称: echo "<select name= 'FirstName'> 而不是以下代码: if(isset($_POST['select'])) { $name = mysqli_real_escape_string($con,$_POST['select']); $fetch = "SELECT * FROM persons WHERE FirstName = '".$name."'"; $result = mysqli_query($con,$fetch);
  3. 另请注意,在where子句中,您应该检查id而不是firstName 使用此更正的代码: if(isset($_POST['FirstName'])) { $name = $_POST['FirstName']; $fetch = "SELECT * FROM persons WHERE id = '".$name."'"; $result = mysqli_query($con,$fetch);

0
投票

它将特定用户的数据显示在同一页面上

你的下拉框看起来像

<select id="c-name" name="name" onChange="reload()">
    <option value="">Select Name</option>
</select>

在Javascript中写reload()函数

function reload()   {
    var val=document.getElementById("c-name").value;
    self.location="yourfilename.php?fname=" + val ;
    //alert(val);
}

在页面上获取fname变量并显示从数据库中获取的该变量的数据

在PHP中

if(isset($_REQUEST['fname'])) {

  //your code

}
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