Question

我需要一些帮助，我有一张名为POSITION的桌子。该表的列是p_no，p_name和p_training_type。 p_name代表位置名称。一个p_name将参加许多类型的培训，例如它可以像这样

p_no | p_name | p_training_type

我想要做的是更新一个p_name例子总经理所有甲肝相同p_name也将change..for例如p_name，用户将搜索他们想要在update1.php页面更新p_name，但是用户不记得位置名称的全名..所以用户将输入MANAGER ..所以结果是。

总经理
HSSE MANAGER

接下来，用户将单击他们想要更新的位置名称。例如，用户单击GENERAL MANAGER。

所以位置名称将被带到另一个页面（update2.php），其中GENERAL MANAGER将以可编辑的形式出现。通过键入GENERAL MANAGER作为ADMIN MANAGER编辑名称n单击更新按钮..它将转到update3.php，此页面中包含更新查询。

对于结果，没有错误显示..但数据stil不更新仍然是GENERAL MANAGER而不是ADMIN MANAGER。我不知道问题出在哪里..以下是更新过程的相关页面..

update1.php

        <form action="update1.php" method="post" name="form1" id="form1">
        <table><tr>
           <th scope="col"><span class="style1"></span>Type:</span></th>
           <th colspan="2"><select name="method">
              <option value="" selected>-Please choose-</option>
              <option value="position">By Position</option>
              <option value="training">By Training Type</option>
        </select></th>
        </tr>
        <tr>
           <th scope="col">Input Query :</th>
           <th><input type="text" id="search" name="search" size="50" /></th>
           <th><input type="submit" name="button" id="button" value="Search" /></th>
        </tr></table>
        </form>

         <?php  $resultName=$_POST['search']; ?>   

<?php

$search =$_POST['search'];
$method =$_POST['method'];

$con = mysql_connect("localhost","root","");
if (!$con)
  {
  die('Could not connect: ' . mysql_error());
  }

mysql_select_db("training", $con);

if($method=="") 
    { 
     }

else if($method=="position"){    
$result = mysql_query("SELECT distinct p_name FROM test WHERE p_name LIKE '%$search%'");
$num_rows = mysql_num_rows($result);
}

else if($method=="training"){    
$result = mysql_query("SELECT distinct p_training_type FROM test WHERE p_training_type LIKE '%$search%'");
$num_rows = mysql_num_rows($result);
} 


echo "<font  size=2>"." Total number of result found for "."<font color='#FF0000' size=2>".$resultName."<font color='#43423F' size=2>"." :"."<font color='#FF0000' size=2>"." $num_rows"."<font color='#43423F' size=2>";
?>
<div
style="border: 0px; width: 620px; height: 400px; overflow: auto;">     
<?php
echo "<table width='600' height='0' border='0' bordercolor='#000000' bgcolor='#FFFFFF'>
<tr>
   <th bgcolor='#000000'><font color='#FFFFFF'>Result </th>
   <th bgcolor='#000000'><font color='#FFFFFF'>Action</th>
</tr>";

if($method!="") 
    {   

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

  if($method=="position"){
   echo "<tr>
            <td height=0 bgcolor='#FFFFFF'>"."<font size=2>".$row['p_name']."</td>
        <td height=0 bgcolor='#FFFFFF'>"."<font size=2><a href='update2.php?p_name=$row[p_name]'><img src='edit.gif' title='Edit' /></a>
         <a href='delete2.php?valid=".$row['p_name']."'><img src='delete.gif' title='Delete' /></a></td>";}

 else if($method=="training"){
  echo "<tr><td height=0 bgcolor='#FFFFFF'>"."<font size=2>"."<input name='p_training_type' width='25' type='text' value='".$row['p_training_type']."'/>"."</td>";}
  }
  }
echo "</table>";

mysql_close($con);

?>

update2.php

<table width="600" border="0">
  <tr>
    <th scope="col">Designation</th>
    <th width="17" scope="col">Action</th>
  </tr>
  <form name="form2"  method="POST" action="update3.php"  enctype="multipart/form-data">
  <tr>
    <td><input name="p_name" type="text" id="p_name" value="<?php echo $row_RecordsetZ['p_name']; ?>" size="20" /></td>
    <td><input type="submit" value="Update" id="update" name="update"/></td>
  </tr>
  </form>

update3.php

<?php
$p_name=$_POST['p_name'];
$con = mysql_connect("localhost","root","");
if (!$con)
  {
  die('Could not connect: ' . mysql_error());
  }

mysql_select_db("training", $con);
{

$sql=mysql_query("UPDATE POSITION SET p_name='$p_name' WHERE p_name='".$p_name."'") or die("Error: " . mysql_error());

    print '<script type="text/javascript">'; 
    print 'alert("The details are successfully insert")'; 
    print '</script>';

    mysql_close($con);

}

    echo "<script language=\"JavaScript\">{                                                                                         
    location.href=\"upate1.php\";                                                                                               
    }</script>";
    header('Content-Type: application/pdf');
?>

Answer 1

您需要通过提交表单

上的隐藏字段来隐藏PID

<form name="form2"  method="POST" action="update3.php"  enctype="multipart/form-data">
  <tr>
     <td><input name="p_id" type="hidden" id="p_id" value="<?php echo $row_RecordsetZ['p_id']; ?>" size="20" /></td>
     <td><input name="p_name" type="text" id="p_name" value="<?php echo $row_RecordsetZ['p_name']; ?>" size="20" /></td>
     <td><input type="submit" value="Update" id="update" name="update"/></td>
  </tr>
</form>

您的提交查询将是

$sql=mysql_query("UPDATE POSITION 
                  SET p_name='$p_name' 
                  WHERE p_name='".$p_name."' AND p_id = '".$_POST['p_id']."'");

从一列更新一个数据名称，所有具有相同名称且在同一列中的数据也将更新

p_no | p_name | p_training_type

1 个答案: