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PHP posting a variable in a variable using mysql


PHP posting a variable in a variable using mysql

By : Oscar Puentes
Date : October 17 2020, 01:08 AM
Does that help If you want to learn about PHP's variable variables, it's in the manual (I linked to it). But you actually don't need it in your case.
Be careful about SQL injection. Your code is vulnerable to it.
code :
$district_num = (int) $_REQUEST['district_num'];

$publish_page_col = "publish_page_{$district_num}";
$publish_page_value = $_REQUEST["publish_page_4{$district_num}"];

$sql = "UPDATE districts SET
        `$publish_page_col` = ?
        WHERE district_num = ?";
$stmt = $pdo->prepare($sql);
$stmt->execute([ $publish_page_value, $district_num ]);


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How do you write "Select (all) From (table) Where posting=$posting except this posting? (Mysql)

How do you write "Select (all) From (table) Where posting=$posting except this posting? (Mysql)


By : Doniv
Date : March 29 2020, 07:55 AM
hope this fix your issue This is the code you need, although it will be better if you have the current post id and just have something like WHERE id != " . (int)$current_post_id . "
code :
$title = mysql_real_escape_string($title);

$sql = "
    SELECT *
    FROM posting
    WHERE
        title LIKE '%" . $title . "%' AND
        detail LIKE '%" . %title . "%' AND
        title != " . $title . "
    ";
$title = mysql_real_escape_string($title);

$sql = "
    SELECT *
    FROM posting
    WHERE
        title LIKE '%" . $title . "%' AND
        detail LIKE '%" . %title . "%' AND
        id != " . (int)$post_id . "
    ";
Posting JavaScript Variable to MySQL using PHP

Posting JavaScript Variable to MySQL using PHP


By : KdoSiOdJinud
Date : March 29 2020, 07:55 AM
I wish this help you Use name parameter for hidden variable and it will be automatically passed to PHP .
code :
<label id="total"></label>

<input type="hidden" name="total" id="nameID"/>
Trouble with image posting and mysql BLOB variable

Trouble with image posting and mysql BLOB variable


By : CSBLACK
Date : March 29 2020, 07:55 AM
To fix this issue Finally I found the answer,tried addslashes function (used for sqlinjection) and surprisingly it worked properly
Is there any way to pass the $scope.variable into the $http.post ? issue # in payload my variable are not posting

Is there any way to pass the $scope.variable into the $http.post ? issue # in payload my variable are not posting


By : alucardpj
Date : March 29 2020, 07:55 AM
this one helps. In HTML i have some customer registration fields , Try this:
code :
var customerInformation = $scope.customer;

$http.post('person/registrationCustomer',customerInformation)
    .then(function (success) {

    }, function (error) {


    });
php - Need echo to evaluate variable twice. Grab variable name from mySQL database, then parse that variable

php - Need echo to evaluate variable twice. Grab variable name from mySQL database, then parse that variable


By : Sudipta Dey
Date : March 29 2020, 07:55 AM
Hope that helps I have on one page: , You need to use eval for this:
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