PHP的MySQL批量记录更新？

Question

I'm trying to update multiple mysql records, with this code:

<strong>Update multiple rows in mysql</strong><br> 

<?php
$host="localhost"; // Host name 
$username=""; // Mysql username 
$password=""; // Mysql password 
$db_name="test"; // Database name 
$tbl_name="test_mysql"; // Table name 

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

$sql="SELECT * FROM $tbl_name";
$result=mysql_query($sql);

// Count table rows 
$count=mysql_num_rows($result);
?>
<table width="500" border="0" cellspacing="1" cellpadding="0">
<form name="form1" method="post" action="">
<tr> 
<td>
<table width="500" border="0" cellspacing="1" cellpadding="0">


<tr>
<td align="center"><strong>Id</strong></td>
<td align="center"><strong>Name</strong></td>
<td align="center"><strong>Lastname</strong></td>
<td align="center"><strong>Email</strong></td>
</tr>
<?php
while($rows=mysql_fetch_array($result)){
?>
<tr>
<td align="center"><? $id[]=$rows['id']; ?><? echo $rows['id']; ?></td>
<td align="center"><input name="name[]" type="text" id="name" value="<? echo $rows['name']; ?>"></td>
<td align="center"><input name="lastname[]" type="text" id="lastname" value="<? echo $rows['lastname']; ?>"></td>
<td align="center"><input name="email[]" type="text" id="email" value="<? echo $rows['email']; ?>"></td>
</tr>
<?php
}
?>
<tr>
<td colspan="4" align="center"><input type="submit" name="Submit" value="Submit"></td>
</tr>
</table>
</td>
</tr>
</form>
</table>
<?php
// Check if button name "Submit" is active, do this 
if($Submit){
   for($i=0;$i<$count;$i++){
     $sql1="UPDATE $tbl_name 
            SET name='$name[$i]', 
                lastname='$lastname[$i]', 
                email='$email[$i]' 
                WHERE id='$id[$i]'";
     $result1=mysql_query($sql1);
   }
}

if($result1){
header("location:update_multiple.php");
?>

It shows records from the DB in input text boxes, but when I change old records with new ones and then submit, nothing happens, page refreshes with old values.

Answer 1

You will need to check if the form is submitted using $_POST['Submit'] . Also you could use a hidden input field to keep track of each row to update:

<input type="hidden" name="id[]" value="<?php echo $row['id']; ?>" />

Then when handling your submission you should do something like this:

if(isset($_POST['Submit'])) {
   $ids = $_POST['id'];
   $names = $_POST['name'];
   $lastnames = $_POST['lastname'];
   $emails = $_POST['email'];
   //
   foreach($ids as $id) {
      // update the record based on the id and supplied data
   }
}

And of course the update process should be executed before you retrieve you rows from the database. In other words the above code should be placed near the top of your script. At least before the:

$sql="SELECT * FROM $tbl_name";
$result=mysql_query($sql);

Answer 2

Submit should define that form if sent

move update code before select because you first view, and just after that update info, so you'll view previous results all the time

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

if($Submit){
for($i=0;$i<$count;$i++){
$sql1="UPDATE $tbl_name SET name='$name[$i]', lastname='$lastname[$i]', email='$email[$i]' WHERE id='$id[$i]'";
$result1=mysql_query($sql1

);
}

$sql="SELECT * FROM $tbl_name";
$result=mysql_query($sql);

// Count table rows 
...

Answer 3

Using if($Submit){ is not the correct way to determinate if a form has been submitted ...

you should use :

if($_SERVER['REQUEST_METHOD'] == "POST")

then process your submitted values.

Note : you should update the DB before showing the values ....

And please make sure you read about SQL Injection

Answer 4

First of all mysql_query() doesn't return anything when you execute an update command. You could use instead mysql_affected_rows to retrieve the number of rows changed by the previous query:

http://php.net/manual/en/function.mysql-affected-rows.php

Then, you should also do some refactoring like moving the update code before displaying the results, otherwise the header location directive won't work if you already sent output to the browser.

Answer 5

did you apply the varaible names in single code?

$sql1="UPDATE $tbl_name SET name='$name[$i]', lastname='$lastname[$i]', email='$email[$i]' WHERE id='$id[$i]'";

change above query to

$sql1="UPDATE $tbl_name SET name='".$name[$i]."', lastname='".$lastname[$i]."', email='".$email[$i]."' WHERE id=$id[$i]";