How to make PHP Update MySQL Table Form display all Rows of Data
Question
I am trying to make this PHP Update Form display all of my row data from the mysql table. While it does work and update my table, I would like it to display each employee rather than having to search up an employee number in order to edit. Basically I would like it to show an array of employees with edit inputs for each, instead of having to search for each one. I am asking this because I have searched vastly across the web and have found no solutions to my appeal.
Any help will be greatly appreciated.
Thank you
PHP Code:
<html>
<head>
<title>Update a Record in MySQL Database</title>
</head>
<body>
<?php
if(isset($_POST['update']))
{
$dbhost = '';
$dbuser = '';
$dbpass = '';
$conn = mysql_connect($dbhost, $dbuser, $dbpass);
if(! $conn )
{
die('Could not connect: ' . mysql_error());
}
$emp_id = $_POST['emp_id'];
$emp_salary = $_POST['emp_salary'];
$sql = "UPDATE employee ".
"SET emp_salary = $emp_salary ".
"WHERE emp_id = $emp_id" ;
mysql_select_db('');
$retval = mysql_query( $sql, $conn );
if(! $retval )
{
die('Could not update data: ' . mysql_error());
}
echo "Updated data successfully\n";
mysql_close($conn);
}
else
{
?>
<form method="post" action="<?php $_PHP_SELF ?>">
<table width="400" border="0" cellspacing="1" cellpadding="2">
<tr>
<td width="100">Employee ID</td>
<td><input name="emp_id" type="text" id="emp_id"></td>
</tr>
<tr>
<td width="100">Employee Salary</td>
<td><input name="emp_salary" type="text" id="emp_salary"></td>
</tr>
<tr>
<td width="100"> </td>
<td> </td>
</tr>
<tr>
<td width="100"> </td>
<td>
<input name="update" type="submit" id="update" value="Update">
</td>
</tr>
</table>
</form>
<?php
}
?>
</body>
</html>
3 answers
solution1
0 2014-02-05 16:18:20
Try to add your sql under form to udpate table. make var_dump to see if you receive values of employer id ad salary.
solution2
0 2014-02-05 16:21:09
Do you want a drop-down menu displaying each employee and then allow the user to enter a new salary for that employee? For that, you could do something like this:
echo "<select name='employee'>";
$query = mysql_query("SELECT * FROM employees");
while($row = mysql_fetch_assoc($query)) {
$id = $row['emp_id'];
echo "<option value='$id'>$id</option>";
}
echo "<input type='text' name='salary'>
echo "</select>";
Forgive me if I misunderstood.
solution3
0 2014-02-05 19:34:34
So you want to display the employee's current salary along with the form asking for the user to input a new one? If so, try something like this:
$employee = $_POST['employee']; // Or however you're retrieving the current employee the user is editing
$query = mysql_query("SELECT * FROM employees WHERE username='$employee'");
echo "<form action='FILENAME.PHP' method='post'>";
while($row = mysql_fetch_assoc($query)) {
$salary = $row['emp_salary'];
echo "Current salary: $salary <br />";
}
echo "<input type='text' name='salary'>";
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