MySQL Table Update with PHP HTML Form isn't Working
Question
I am trying to update mySQL table using a form with PHP. Currently I have all the code set up but when I am updating my table data, age, it sets all the ages in the table to '0'. I am not sure why but any guidance would be strongly appreciated. Thanks.
Kelsey
<?php
$hostname = "---------";//host name
$dbname = "-------";//database name
$username = "-------------";//username you use to login to php my admin
$password = "--------";//password you use to login
//CONNECTION OBJECT
//This Keeps the Connection to the Databade
$conn = new MySQLi($hostname, $username, $password, $dbname) or die('Can not connect to database')
?>
<!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Transitional//EN" "http://www.w3.org/TR/xhtml1/DTD/xhtml1-transitional.dtd">
<html xmlns="http://www.w3.org/1999/xhtml">
<head>
<meta http-equiv="Content-Type" content="text/html; charset=UTF-8" />
<title>Untitled Document</title>
</head>
<body>
<?php
$id=$_GET['FirstName'];
//Create a query
$sql = "SELECT * FROM Persons";
//submit the query and capture the result
$result = $conn->query($sql) or die(mysql_error());
$query=getenv(QUERY_STRING);
parse_str($query);
?>
<h2>Update Record <?php echo $sql;?></h2>
<form action="" method="post">
<?php
while ($row = $result->fetch_assoc()) {?>
<table border="0" cellspacing="10">
<tr>
<td>age:</td> <td><input type="text" name="Age" value="<?php echo $row['Age']; ?>"></td>
</tr>
<tr>
<td><INPUT TYPE="Submit" VALUE="Update the Record" NAME="Submit"></td>
</tr>
</table>
<?php }
?>
</form>
<?php
if(isset($_POST['Submit'])){//if the submit button is clicked
$sql="UPDATE Persons SET Age='".$_POST['Age']."'";
$conn->query($sql) or die("Cannot update");//update or error
}
?>
</body>
</html>
2 answers
solution1
1 2014-02-04 16:40:02
The UPDATE query as its written right now updates the entire Persons table, not an individual record.
UPDATE Persons SET Age=15 WHERE id = 5 will only update one record as apposed to the entire table's values.
Also, it's not good (aka massive security risk) to put raw post values directly into an SQL string. You should always sanity check your values before putting them into a database query.
solution2
0 ACCPTED 2014-02-04 16:56:09
Not a good way to add the POST inside the statement itself.
Try this:
<?php
if(isset($_POST['Submit'])){//if the submit button is clicked
$age = $_POST['Age'];
$sql="UPDATE Persons SET Age='".$age."' WHERE firstname ='".$id."'";
$conn->query($sql) or die("Cannot update");//update or error
}
?>
Your page is confusing. You didn't even output the result correctly. Try this:
<?php
$hostname = "---------";//host name
$dbname = "-------";//database name
$username = "-------------";//username you use to login to php my admin
$password = "--------";//password you use to login
//CONNECTION OBJECT
//This Keeps the Connection to the Databade
$conn = new MySQLi($hostname, $username, $password, $dbname) or die('Can not connect to database')
?>
<!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Transitional//EN" "http://www.w3.org/TR/xhtml1/DTD/xhtml1-transitional.dtd">
<html xmlns="http://www.w3.org/1999/xhtml">
<head>
<meta http-equiv="Content-Type" content="text/html; charset=UTF-8" />
<title>Untitled Document</title>
</head>
<body>
<?php
//$id=$_GET['FirstName'];
//Create a query
$sql = "SELECT * FROM Persons";
//submit the query and capture the result
$result = $conn->query($sql) or die(mysql_error());
$query=getenv(QUERY_STRING);
parse_str($query);
?>
<h2>Update Record <?php echo $sql;?></h2>
<form action="" method="post" enctype='multipart/form-data'>
<?php
while ($row = $result->fetch_assoc()) {?>
<table border="0" cellspacing="10">
<tr>
<td>Age:</td>
<td><?php echo $row['Age'];?></td>
<td><?php echo $row['FirstName'];?></td>
<td><input type="text" name="Age"></td>
<td><input type=hidden" name="firstName" value="<?php echo $row['FirstName'];?>"></td>
</tr>
<tr>
<td><INPUT TYPE="Submit" VALUE="Update the Record" NAME="Submit"></td>
</tr>
</table>
<?php }
?>
</form>
<?php
if(isset($_POST['Submit'])){//if the submit button is clicked
$age = $_POST['Age'];
$id = $_POST['firstName'];
$sql="UPDATE Persons SET Age='".$age."' WHERE firstname ='".$id."'";
$conn->query($sql) or die("Cannot update");//update or error
}
?>
</body>
</html>
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