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Closed 7 years ago.
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Can anyone see whats wrong with this syntax? I've consulted the wordpress codex page and the syntax seems to be correct..
$result = $wpdb->query($wpdb->prepare( "SELECT * FROM {$wpdb->prefix}users WHERE username = %s AND password = %s, array( $username, $password )"));
Thanks
try:
$result = $wpdb->query($wpdb->prepare( "SELECT * FROM {$wpdb->prefix}users WHERE username = %s AND password = %s", array( $username, $password )));
your quotes are incorrect.
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Closed 3 years ago.
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I am trying to prepare an insertion inside the database with php, but i can't find any suggestions on how to use the syntax, this is what I've got so far, could someone help me with this please?
$last_id = mysqli_insert_id($conn);
$query = $conn->prepare("INSERT INTO `table`(name, surname) VALUES (?,?) , (?,?)");
$query->bind_param("ss", ($nameOne, $last_id), ($nametWO, $last_id));
the binding should be
$query->bind_param("ssss", $nameOne, $last_id, $nametWO, $last_id);
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Closed 6 years ago.
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Hello I have little problem with this:
require_once("dbconnect.php");
mysql_query('set names utf8');
$query = ('DELETE FROM `gallery` WHERE `id` = 16') or die("failed");
It not delete record from DB. Can you help me? Thanks. dbconnect.php is correct.
All you did was declare a variable $query, but you did not run it. You have to run it is $query = mysql_query('DELETE FROM `gallery` WHERE `id` = 16')
Edit: Having said that, do not use mysql_ functions. They allow for SQL injection, which is very dangerous. Instead it is better to use
$mysqli = new mysqli(HOST, USERNAME, PASSWORD, DATABASE);
$query = $mysqli->query('DELETE FROM `gallery` WHERE `id` = 16');
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Closed 6 years ago.
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I am trying to create a query which will update each person's profile views when a member accesses their profile. For some reason it does not update, I am sure it would have worked. Here is my lines:
<?php
$name = $_REQUEST['username'];
$profile = mysql_query("SELECT * FROM users WHERE user_name='$name'");
$uprofile = mysql_fetch_assoc($profile);
$username = $uprofile['user_name'];
$profilev = $uprofile['profile_views'];
mysql_query("UPDATE users SET profile_views +1 WHERE user_name='$name'");
?>
So, On my profile, It displays their username successfully, and it displays the default value of the database for profile_views which ofcourse, is 0. so it is reading correctly, I am just having issues updating the users profile_views.
You have a mistake try
"UPDATE users
SET profile_views = profile_views +1
WHERE user_name='$name'"
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Closed 8 years ago.
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Can someone tell me what is wrong with that statement cause I do not see any mistake?
$PrevReleaseModel = $con->prepare("SELECT * FROM model WHERE model_name=:model_name AND model_release:model_release");
$PrevReleaseModel->bindParam('model_name',$model_name);
$PrevReleaseModel->bindParam('model_release',$model_release);
$PrevReleaseModel->execute(array('model_name'=>$model_name,'model_release'=>$model_release));
I am really confused.
WHERE model_name=:model_name AND model_release = :model_release
You missed a equal sign in the last condition.
You may try this out:-
$PrevReleaseModel = $con->prepare('SELECT * FROM model
WHERE model_name = :model_name AND
model_releas = :model_releas'
);
$PrevReleaseModel->bindParam(':model_name', $model_name, PDO::PARAM_STR);
$PrevReleaseModel->bindParam(':model_releas', $model_release, PDO::PARAM_STR);
$PrevReleaseModel->execute();
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Closed 8 years ago.
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it says unexpected "="... What should I rewrite? Thanks
$result = mysql_query("SELECT * FROM soubory, users
WHERE id='".$id."'" AND soubory.users_id = users.id );
Remove the second " after $id."'
$result = mysql_query("SELECT * FROM soubory, users WHERE soubory.users_id='".$id."' AND soubory.users_id = users.id");