[英]Updating Form with MYSQL data
我是PHP新手,想創建一個表單,在該表單中,用戶將數據插入到表單中(有效),然后將其存儲在MYSQL DB上(有效),現在必須顯示數據,然后必須能夠進行修改某些記錄,現在我有顯示記錄的部分以及“編輯”按鈕,但是由於同一條記錄不斷出現,某處出現了問題,所以我猜我的代碼有問題:(
請幫忙:
這是index.php代碼:
<?php
include('dbinfo.php');
$sql="SELECT * FROM stats";
$result = mysql_query($sql, $db) or die (mysql_error());
$pageTitle = "My Stats Database";
include("header.php");
print <<<HERE
<h2> My Contacts</h2>
Select a Record to update <a href="addstat.php"> add new stat</a>.
<table id="home">
HERE;
while ($row=mysql_fetch_array($result)){
$id=$row["id"];
$type=$row["type"];
$depthead=$row["depthead"];
$person=$row["person"];
$descr=$row["descr"];
$recdate=$row["recdate"];
$tolog=$row["tolog"];
$senttorev=$row["senttorev"];
$recfromrev=$row["recfromrev"];
print <<<HERE
<tr>
<td>
<form method="POST" action="updateform.php">
<input type="hidden" name="sel_record" value="$id">
<input type="submit" name="update" value=" Edit " </form>
</td>
<td><strong> Description: </strong>$descr,<p> <strong>Type: </strong>$type</p> <p><strong> Department Head: </strong>$depthead</p>
<strong> Test Analyst: </strong> $person<br/></td>
HERE;
}
print "</tr></table></body></html>";
?>
然后這是我的更新updateform.php腳本:
<?php
include("dbinfo.php");
$sel_record = $_POST['sel_record'];
//$sel_record = (isset($_POST['sel_record'])) ? $_POST['sel_record'] : '';
$sql = "SELECT * FROM stats WHERE id = 'sel_record'";
//execute sql query and get result
$result = mysql_query($sql, $db) or die (mysql_error());
if (!$result) {
print "<h1> Something went wrong!</h1>";
} else
{ //begin while loop
while ($record = mysql_fetch_array($result, MYSQL_ASSOC)){
$id = $record["id"];
$type = $record['type'];
$depthead = $record['depthead'];
$person = $record["person"];
$descr = $record["descr"];
$recdate = $record["recdate"];
$tolog = $record["tolog"];
$senttorev = $record["senttorev"];
$recfromrev = $record["recfromrev"];
}
}
//end while loop
$pagetitle = "Edit Stat";
include ("header.php");
print <<<HERE
<h2> Modify this Stat</h2>
<p> Change the values in the boxes and click "Modify Record" button </p>
<form id="stat" method="POST" action="update.php">
<input type="hidden" name="id" value="$id">
<div>
<label for="type">Type*:</label>
<input type="text" name="type" id="type" value="$type">
</div>
<p>
</p>
<div>
<label for = "depthead" >Department Head*:</label>
<input type = "text" name = "depthead" id = "depthead" value = "$depthead">
</div>
<p>
</p>
<div>
<label for="person">Test Analyst*:</label>
<input type="text" name="person" id="person" value="$person">
</div>
<p>
</p>
<div>
<label for="descr">Description*:</label>
<input type="text" name="descr" id="descr" value="$descr">
</div>
<p>
</p>
<div>
<label for="recdate">Date Received*:</label>
<input type="text" name="recdate" id="recdate" value="$recdate">
</div>
<p>
</p>
<div>
<label for="tolog">Date to log*:</label>
<input type="text" name="tolog" id="tolog" value="$tolog">
</div>
<p>
</p>
<div>
<label for="senttorev">Sent to Rev:</label>
<input type="text" name="senttorev" id="senttorev" value="$senttorev">
</div>
<p>
</p>
<div>
<label for="recfromrev">Received from Rev*:</label>
<input type="text" name="recfromrev" id="recfromrev" value="$recfromrev">
</div>
<p>
</p>
<div id="mySubmit">
<input type="submit" name="submit" value="Modify Record">
</div>
</form>
HERE;
?>
然后,mysql的實際更新具有一個update.php腳本:
<?php
include "dbinfo.php";
$id = $_POST['id'];
$type = $_POST['type'];
$depthead = $_POST['depthead'];
$person = $_POST['person'];
$descr=$_POST['descr'];
$recdate=$_POST['recdate'];
$tolog=$_POST['tolog'];
$senttorev=$_POST['senttorev'];
$recfromrev=$_POST['recfromrev'];
$sql="UPDATE stats SET
depthead='$depthead',
person='$person',
descr='$descr',
recdate='$recdate',
tolog='$tolog',
senttorev='$senttorev',
recfromrev='$recfromrev'
WHERE id='$id'";
$result=mysql_query($sql) or die (mysql_error());
print "<html><head><title>Update Results</titlel></head><body>";
include "header.php";
print <<<HERE
<h1>The new Record looks like this: </h1>
<td>
<p><strong>Type: </strong>$type</p>
<p><strong>Department Head: </strong>$depthead</p>
<p><strong>Test Analyst: </strong> $person</p>
<p><strong>Description: </strong>$descr</p>
<p><strong>Received Date:</strong>$recdate</p>
<p><strong>Date to Log:</strong>$tolog</p>
<p><strong>Sent to rev:</strong>$senttorev</p>
<p><strong>Received from Rev:</strong>$recfromrev</p>
<br/>
HERE;
有人可以告訴我為什么只保留其中一條記錄與從index.php頁中選擇哪一條無關緊要。 出於某種原因,我認為這是我的$sel_record
變量,但我不確定並用盡了Ideas。
先感謝您..
這是您在updateform.php中的問題:
$sql = "SELECT * FROM stats WHERE id = 'sel_record'";
應該是:
$sql = "SELECT * FROM stats WHERE id = $sel_record";
您錯過了$符號來調用變量,並且不需要在ID兩端加上引號。
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