当我在 php 页面上使用表单更新 Mysql 表时,当我点击提交时页面不会自动刷新。我必须点击刷新才能看到更新的行。我试过了,但这也没有奏效,没有任何好主意?
<META HTTP-EQUIV="CACHE-CONTROL" CONTENT="NO-CACHE">
<META HTTP-EQUIV="PRAGMA" CONTENT="NO-CACHE">
///////////////////////Edited Bottom////////////////////////////////////////////////////////
<?php
include ("../header.php");
include ("../header2.php");
?>
<?php
include('../config.php');
include ('../ac.thumbs.php');
if ($_GET['sil']) {
$sil = htmlspecialchars(stripslashes($_GET['sil']));
mysql_query("DELETE FROM CloseOut WHERE itemNO = $sil limit 1");
echo "<h1>$sil item nolu kayit silinmistir</h1><br>";
}
$edit = htmlspecialchars(stripslashes($_GET['edit']));
$data = mysql_query("SELECT * FROM CloseOut WHERE itemNO=$edit") or die(mysql_error());
//Puts it into an array
while($info = mysql_fetch_array( $data ))
{ echo "
<center>
<a href='"edit.php?sil=$info[itemNO]'"onclick='"return confirm('Do you really want to erase this entry?')'">[ Delete ]</a><br>
<a href=../../large.php?f=upload/".$info['resim'], $thumblarge .">
<img src=../../small.php?f=upload/".$info['resim'], $thumbsmall .">
</a>";
$editresim = $info[resim];
$editisim = $info[isim];
$editalan1 = $info[alan1];
$editalan2 = $info[alan2];
$editalan3 = $info[alan3];
$editalan4 = $info[alan4];
}
//Resimlerin yuklenecegin yer
$target = "../../upload/";
//////////////////////////////////////////////////////////////////////
///////////////////////////////Resim1/////////////////////////////////
//////////////////////////////////////////////////////////////////////
$target1 = $target . basename( $_FILES['resim1']['name']) ;
//Formdan gelen bilgileri almasi icin
$resim=($_FILES['resim1']['name']);
$isim=$_POST['isim1'];
$alan1=$_POST['alan11'];
$alan2=$_POST['alan21'];
$alan3=$_POST['alan31'];
$alan4=$_POST['alan41'];
/////////////Formdaki bos alanlara bos bilgi kayit etmemesi icin////////////////////////
$updates = array();
if (!empty($resim))
$updates[] = 'resim="'.mysql_real_escape_string($resim).'"';
if (!empty($isim))
$updates[] = 'isim="'.mysql_real_escape_string($isim).'"';
if (!empty($alan1))
$updates[] = 'alan1="'.mysql_real_escape_string($alan1).'"';
if (!empty($alan2))
$updates[] = 'alan2="'.mysql_real_escape_string($alan2).'"';
if (!empty($alan3))
$updates[] = 'alan3="'.mysql_real_escape_string($alan3).'"';
if (!empty($alan4))
$updates[] = 'alan4="'.mysql_real_escape_string($alan4).'"';
$updates = implode(', ', $updates);
/////////////////////////////////////////////////////////////////////////////////////////
//update etmesi icin
if(move_uploaded_file($_FILES['resim1']['tmp_name'], $target1))
{
mysql_query("UPDATE CloseOut SET $updates WHERE itemNO='$edit' ");
echo "Tebrikler, ". basename( $_FILES['resim1']['name']). " isimli dosya yuklendi ve database'e basariyla islendi!<br>";
}
?>
<br>
<center>
<table width="900" border="5" style="background-color:white; "bordercolor="#000000">
<tr>
<td width="120" height="50" bgcolor= "545454"><b>RESIM</b></td>
<td width="120" height="50" bgcolor= "545454"><b>NAME</b></td>
<td width="120" height="50" bgcolor= "545454"><b>CATAGORY</b></td>
<td width="120" height="50" bgcolor= "545454"><b>QUANTITY</b></td>
<td width="120" height="50" bgcolor= "545454"><b>LOCATION</b></td>
<td width="120" height="50" bgcolor= "545454"><b>PRICE</b></td>
</tr>
<tr>
<form enctype="multipart/form-data" method="post" action="">
<td><input type="file" name="resim1" value="<?php echo $editresim;?>"> </td>
<td><input type="text" name="isim1" value="<?php echo $editisim;?>"> </td>
<td><input type="text" name="alan11" value="<?php echo $editalan1;?>"> </td>
<td><input type="text" name="alan21" value="<?php echo $editalan2;?>"> </td>
<td><input type="text" name="alan31" value="<?php echo $editalan3;?>"></td>
<td><input type="text" name="alan41" value="<?php echo $editalan4;?>"></td>
</tr>
<td colspan="6"><center><input type="submit"></center></td>
</form>
</table>
</center>
你现在能看到吗? $editresim
值为 $info[resim]
。
提交表单时,$info[resim]
是当前数据库值。数据库在下面/之后更新了。
提交时,您仍会看到提交的值实际放入数据库之前的先前数据。因此,您需要重新打开该页面才能查看更新的数据。
修复它的一种方法是,您应该在SELECT * FROM CloseOut WHERE itemNO=$edit
之前将代码以更新数据移动到顶部,以便查询将为您提供最新/更新的数据。
很抱歉没有提供示例代码,但恐怕我会犯一些小错误,这可能会增加混乱。如果你尝试自己解决它,实际上会更好。
祝你好运!
ps:你的逻辑
Get data from db
Put to data to VAR
If form is submitted:
Update data in DB
Show VAR // This is *not* the latest submitted data
问题是您在获取信息后正在更新数据库。
把这部分代码
//Resimlerin yuklenecegin yer
$target = "../../upload/";
//////////////////////////////////////////////////////////////////////
///////////////////////////////Resim1/////////////////////////////////
//////////////////////////////////////////////////////////////////////
$target1 = $target . basename( $_FILES['resim1']['name']) ;
//Formdan gelen bilgileri almasi icin
$resim=($_FILES['resim1']['name']);
$isim=$_POST['isim1'];
$alan1=$_POST['alan11'];
$alan2=$_POST['alan21'];
$alan3=$_POST['alan31'];
$alan4=$_POST['alan41'];
/////////////Formdaki bos alanlara bos bilgi kayit etmemesi icin////////////////////////
$updates = array();
if (!empty($resim))
$updates[] = 'resim="'.mysql_real_escape_string($resim).'"';
if (!empty($isim))
$updates[] = 'isim="'.mysql_real_escape_string($isim).'"';
if (!empty($alan1))
$updates[] = 'alan1="'.mysql_real_escape_string($alan1).'"';
if (!empty($alan2))
$updates[] = 'alan2="'.mysql_real_escape_string($alan2).'"';
if (!empty($alan3))
$updates[] = 'alan3="'.mysql_real_escape_string($alan3).'"';
if (!empty($alan4))
$updates[] = 'alan4="'.mysql_real_escape_string($alan4).'"';
$updates = implode(', ', $updates);
/////////////////////////////////////////////////////////////////////////////////////////
//update etmesi icin
if(move_uploaded_file($_FILES['resim1']['tmp_name'], $target1))
{
mysql_query("UPDATE CloseOut SET $updates WHERE itemNO='$edit' ");
echo "Tebrikler, ". basename( $_FILES['resim1']['name']). " isimli dosya yuklendi ve database'e basariyla islendi!<br>";
}
就在之前:
$edit = htmlspecialchars(stripslashes($_GET['edit']));