Php Mysql Update

Question

I have some code that I am trying to make into a news feed for my front page. So far I cant get the Update to work. Ps feel free to clean up code and make for practical as this is version 1 I haven't cleaned code up. Thanks Joshua

Update Part

    if(isset($_POST['edit']))
    {
    $checkbox = $_POST['checkbox'];

    for($i=0;$i<count($_POST['checkbox']);$i++){

    $edit_id = $checkbox[$i];

        echo "<textarea name='editsavetext' id='editsavetext' rows='4' cols='50'>$edit_id</textarea>";
    }};

    if(isset($_POST['editsave'])){

    echo $editsavetext;

    $editsavetext = $_POST['editsavetext'];

    $sql='UPDATE $tbl_name SET home_show = $editsavetext WHERE home_show = $edit_id';

    $result = mysql_query($sql);

    if($result){
echo "<meta http-equiv='refresh' content='0; url=http://****' />";}};

For Rich Bradshaw

Well at the moment the code doesnt work because i need the system to remember what checkbox was ticks and what is being edited using $edit_id because it doesnt remember the results when being saved so it doesnt effect anything as such.

if(isset($_POST['edit']))
{
$checkbox = $_POST['checkbox'];

for($i=0;$i<count($_POST['checkbox']);$i++){

$edit_id = $checkbox[$i];

    echo "<textarea name='editsavetext' id='editsavetext' rows='4' cols='50'>$edit_id</textarea>";
}};

if(isset($_POST['editsave'])){

echo $editsavetext;

$editsavetext = $_POST['editsavetext'];

$sql="UPDATE `$tbl_name` SET home_show = '$editsavetext' WHERE home_show=':edit_id'";

$result = mysql_query($sql);

if($result){echo "<meta http-equiv='refresh' content='0; url=http://****' />";}};

If you change your query to $sql="UPDATE $tbl_name SET home_show = '$editsavetext' WHERE home_show = '$edit_id'";, will it work? — Logan Wayne
– Logan Wayne, Commented Oct 1, 2014 at 9:03
Note: the mysql_* functions are deprecated and will be removed from future versions of PHP. You should not write new code using them. Use mysqli_* or PDO instead. — Gerald Schneider
– Gerald Schneider, Commented Oct 1, 2014 at 9:04

Rich Bradshaw · Accepted Answer · 2014-10-01 09:11:27Z

1

Without digging through all of that,

'UPDATE $tbl_name SET home_show = $editsavetext WHERE home_show = $edit_id'

should be

"UPDATE `$tbl_name` SET home_show = '$editsavetext' WHERE home_show = $edit_id"

if you had code like:

$a = "hi";

echo 'I say $a';

it would write out I say $a.

if you had

echo "I say $a";

it would write I say hi.

Also, text needs to be in speechmarks in SQL queries.

Then you need to look up SQL Injection, as this code is not safe to use without protecting against it. Finally, why are you using mysql_query? Read the docs on it: http://php.net/manual/en/function.mysql-query.php - note that it's deprecated.

enter image description here

edited Oct 1, 2014 at 9:11

answered Oct 1, 2014 at 9:03

Rich Bradshaw

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11 Comments

Logan Wayne Over a year ago

Oops.. home_show='$edit_id'. I understand, you eagerly wants to post your answer. ;)

Rich Bradshaw Over a year ago

I wouldn't bother doing that as it's an integer.

Rich Bradshaw Over a year ago

At this point you need to check through the rest of the code and make sure you haven't made the same/similar mistakes elsewhere. Reduce your code to just the sql part, does that work? If so, this problem is solved, but you have more problems.

user3818340 Over a year ago

I worked out the problem is inserting the output $edit_id into it but i havn't fixed it yet

Rich Bradshaw Over a year ago

I'd assumed that $edit_id was an integer, if not, then wrap that in speechmarks too.

|

Kamehameha · Accepted Answer · 2014-10-01 09:10:48Z

0

Your update statement seems to be wrong.
You are using single quotes and using $edit_id inside it. Use double quotes for that.
Try this (Assuming that home_show are ints.)-

$sql="UPDATE $tbl_name SET home_show = $editsavetext WHERE home_show = $edit_id";
     ^                                                                         ^

Assuming home_show column is varchar

$sql="UPDATE $tbl_name SET home_show = '$editsavetext' WHERE home_show = '$edit_id'";
     ^                                 ^             ^                   ^        ^^

edited Oct 1, 2014 at 9:10

answered Oct 1, 2014 at 9:04

Kamehameha

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