Question

如何使用php

更新mysql中的表单选择2行值

我有这个链接

floorsedit.php?id=Building1&floorno=1

当我访问此链接时它显示空白请告诉我为什么它显示空白以及为什么它不更新我认为有一个问题就是为什么它没有显示任何内容

$sql="SELECT * FROM $tbl_name where buildingname='".$id."' and floorno='".$floorno."' ";

请帮我解决这个问题

感谢 ...

这是主页

floorsedit.php

    $host="localhost"; // Host name 
    $username="root"; // Mysql username 
    $password=""; // Mysql password 
    $db_name="building"; // Database name 
    $tbl_name="floors"; // Table name

    // Connect to server and select database.
    mysql_connect("$host", "$username", "$password")or die("cannot connect"); 
    mysql_select_db("$db_name")or die("cannot select DB");

    // get value of id that sent from address bar
     $id = $_GET['id']; 

     $floorno = $_GET['$floorno'];

    // Retrieve data from database 
    $sql="SELECT * FROM $tbl_name where buildingname='".$id."' and floorno='".$floorno."' ";
    $result=mysql_query($sql);
    $rows=mysql_fetch_array($result);
    ?>
    <?php
    // close connection 
    mysql_close();
    ?>

and this is a form
<form name="form1" method="post" action="update_ac.php" class="registration_form">

  <fieldset>
    <legend>New Floor </legend>
<input name="id" type="hidden" id="buildingname" value="<? echo $rows['buildingname']; ?>">

    <p>Create A New Floor of Building <span class="style4" style="background:#EAEAEA none repeat scroll 0 0;line-height:1;margin-left:410px;;padding:9px 9px;">Please Fill the All Info </span> </p>

    <div class="elements">
      <label for="buildingname">Building Name  :</label>
      <input type="text" id="buildingname" name="buildingname" value="<? echo $rows['buildingname']; ?>" size="25" />
       <label2 for="floorno">Floor No   :</label2>
      <input type="text" id="floorno" name="floorno" value="<? echo $rows['floorno']; ?>" size="25" />
</div>
    <div class="elements">
      <label for="flatno">Floor No of Flats  :</label>
      <input type="text" id="flatno" name="flatno" value="<? echo $rows['flatno']; ?>" size="25" />

    </div>

    </p >

    <div class="submit">
<input type="submit" name="Submit" value="Submit">
    </div>
  </fieldset>
</form>

update_ac.php

<?php
$host="localhost"; // Host name 
$username="root"; // Mysql username 
$password=""; // Mysql password 
$db_name="building"; // Database name 
$tbl_name="floors"; // Table name 

// Connect to server and select database.
mysql_connect("$host", "$username", "$password")or die("cannot connect"); 
mysql_select_db("$db_name")or die("cannot select DB");

// update data in mysql database 
$sql="UPDATE $tbl_name SET buildingname='$buildingname', floorno='$floorno', flatno='$flatno' WHERE buildingname='$id' AND floorno='$floorno'";
$result=mysql_query($sql);

// if successfully updated. 
if($result){
echo "Successful";
echo "<BR>";
echo "<a href='vieweditfloor.php'>View result</a>";
}

else {
echo "ERROR";
}

?>

Answer 1

在floorsedit.php中，使用

$floorno = $_GET['floorno'];

而不是

$floorno = $_GET['$floorno'];

我认为问题出在这一行。希望它有所帮助...

如何在mysql中更新表单选择2行值

1 个答案: