如何显示表中的值，修改它们，再次保存它们

I am trying to display values from table and want to modify the values on form submit.

here is my code:

<?php
$sql = mysql_query("SELECT * FROM TABLENAME");

//Get and store values into new variable

while ($row = mysql_fetch_array($sql)) {
    $name[]   = $row['name'];
    $status[] = $row['status'];

}

//run for loop to identify values

for ($i = 0; $i < count($name); $i++) {
        echo "<tr> ";
        echo "<td>" . $name[$i] . "</td>";
        echo "<td><input  name='offer_new_name' size='5' type='text' value='$status[$i]' /></td>";

}
echo '<form action="" method="post">';
echo '<input type="submit" value="Save New Values" name="save_new">';
echo "</form>";

if (isset($_REQUEST['save_new'])) {
    for ($i = 0; $i < count($name); $i++) {
        echo "name:$name[$i]<br>";
        echo "status:$status[$i]";
    }
}

When I click Save New Values , it displays the old values ,, taken from db. Please guide me how to save

echo "<td><input  name='new_status' size='5' type='text' value='$status[$i]' /></td>";

into new variable for each name and save them into the same table again.

Please guide me if my approach is wrong towards this requirement.

Update1:

table structure:

id       name:               status:
auto     name1                  0
auto     name2                  1
auto     name3                  0

Where this runs,

 while ($row = mysql_fetch_array($sql)) {
        $name[]   = $row['name'];
        $status[] = $row['status'];

    }

values are stored like this

 $name[0]  
  $status[0] 
  $name[1]  
  $status[1]
  $name[2]  
  $status[2]

I can run update query like this

UPDATE TABLE set status=$status[0] where name= $name[0]

First, the following line should be in the <form></form>

echo "<td><input  name='new_status' size='5' type='text' value='$status[$i]' /></td>";

Second, this code block should be on top of your script:

if(isset($_REQUEST['save_new'])){
  //Your update query
}

In this code block you need an update query, which updates your database e.g

UPDATE TABLENAME SET a = new_value WHERE b = c;

This is the variable you can access the entered value;

$_POST['new_status'];

Lastly I can strongly recommend to use prepared statements. Otherwise your code will be open for SQL-Injections.

Pseudo code:

<?php
//if the submit button is pressed, update the Database
if (isset($_REQUEST['save_new'])) {
    //Update your Database
}

//get the new values from Database
$sql = mysql_query("SELECT * FROM TABLENAME");

//store values into new variable
$name = array();
$status = array();
$id = array();
while ($row = mysql_fetch_array($sql)) {
    $name[]   = $row['name'];
    $status[] = $row['status'];
    $id[] = $row['id'];
}

//create HTML-From
echo '<form action="" method="post">';
for ($i = 0; $i < count($name); $i++) {
    echo "<tr> ";
    echo "<td>" . $name[$i] . "</td>";
    echo "<td><input  name='new_status' size='5' type='text' value='$status[$i]' /></td>";
}
echo '<input type="submit" value="Save New Values" name="save_new">';
echo "</form>";
?>

This 'new id' can be saved as a hidden input in your script:

$indecipherable_id = (int)$id * 7258289382234; //Only a example --> use a more complex algorithm!

Before your update query you can decrypt the id and select the row.