Fetch assoc php ошибка

Почему выскакивает такая ошибка?

Fatal error: Call to a member function fetch_assoc() on a non-object in Z:homeAnastasiyawwwsearch.php on line 22

Хочу сделать поиск по сайту, с помощью БД.
Создала БД «anastasiya» (в которой и буду совершать поиск) с таблицей «content».

Вот код:

    $areaywords = explode(" ", $words);
    print_r($areaywords);
    foreach ($areaywords as $key => $value){
        if (isset($areaywords[$key - 1]))
        $query_search .= ' OR ';
        $query_search .='`title` LIKE "%'.$value.'%" OR `text_article` LIKE "%'.$value.'%"';
    }

    $query = "SELECT * FROM content WHERE $query_search";
    echo $query;
    $mysqli = new mysqli("localhost", "root", "", "anastasiya");
    $result_set = $mysqli->query($query);
    $mysqli -> close();

    $i = 0;
    while ($row = $result_set->fetch_assoc()){
        $results[$i] = $row;
        $i++;
    }
    return $results;

}
if (isset ($_POST['bcearch'])){
    $words = $_POST['words'];
    $results = search($words);
}

Ругается на строку:

while ($row = $result_set->fetch_assoc()){

задан 6 окт 2015 в 12:42

here is my code for the query in php:

$query3 = mysql_query("SELECT * FROM area_of_work") or die('Invalid query:'. mysql_error());

while($query3 = mysql_fetch_assoc($query3)){
    $flag = true;
    foreach($listOfUserAreas as $obj){
        if($obj->areaId == $query3['id']){
            $flag = false;
            break;
        }  
    }
    if($flag){
        $areaObj = new AreaInfo();
        $areaObj->areaId = $query3['id'];
        $areaObj->areaName = $query3['areaOfWork'];
        $areaObj->areaTableName = $query3['tableName'];
        array_push($listOfUnusedAreas, $areaObj);
    }
}

and i am getting this error:

Warning: mysql_fetch_assoc() expects parameter 1 to be resource, array given in C:xampphtdocsjob-skillsphpfunctionsuser_unused_area_list.php on line 30
[]

can’t see something wrong.
Thanks

don’t use the same variables.. try like this

$query3 = mysql_query("SELECT * FROM area_of_work") or die('Invalid query:'. mysql_error());

while($result = mysql_fetch_assoc($query3)){
.....//Your code
}

answered Oct 4, 2012 at 11:52

In this line:

while($query3 = mysql_fetch_assoc($query3))

you are reassigning $query3 with results of mysql_fetch_assoc

Should be something like this:

while($queryRes = mysql_fetch_assoc($query3))

answered Oct 4, 2012 at 11:51

Your problem is here:

$query3 = mysql_query("SELECT * FROM area_of_work") or die('Invalid query:'. mysql_error());
while ($query3 = mysql_fetch_assoc($query3)) {

It should be like this:

$q = mysql_query("SELECT * FROM area_of_work") or die('Invalid query:'. mysql_error());
while ($query3 = mysql_fetch_assoc($q)) {

answered Oct 4, 2012 at 11:52

Rename variable $query3 to $query or something different from variable used in your while loop.

What you do on line: while($query3 = mysql_fetch_assoc($query3)) is that you assign result of mysql_fetch_assoc to $query3 (it’s resource at this point) on the first iteration and on the second you pass $query3 (an array already) to mysql_fetch_assoc.

Note: mysql_* functions are deprecated. Use PDO or mysqli_*

answered Oct 4, 2012 at 11:51

Your first loop will work ok, but your second will fail:

while($query3 = mysql_fetch_assoc($query3)){

After the first loop has executed, you are assigning the result array to the resource, effectingly overwriting the resource.

Replace your line like this:

while($listOfUserAreas = mysql_fetch_assoc($query3)){

answered Oct 4, 2012 at 11:51

This is wrong:

while($query3 = mysql_fetch_assoc($query3)){

You are using the same variable to read from, an to assign to.

You can fix it by replacing the aforementioned line, for this one:

$result = &$query3;
while($query3 = mysql_fetch_assoc($result)){

just changing those lines, the rest of your code will work.

answered Oct 4, 2012 at 11:51

You’re trying to assign associate array to resource

while($query3 = mysql_fetch_assoc($query3))

which is wrong… try with different variable
while($Result = mysql_fetch_assoc($query3))

answered Oct 4, 2012 at 11:52

while($query4 = mysql_fetch_assoc($query3)){

change name of fetch variable

answered Oct 4, 2012 at 11:55

С тем, что запрос не правильный — » вокруг имени таблицы не на месте. И в итоге у вас вернулся FALSE, а не mysqli_result.

К тому же, лучше использовать PDO.

21 июн. 2023, в 23:43

5000 руб./за проект

21 июн. 2023, в 23:34

10000 руб./за проект

21 июн. 2023, в 22:52

600 руб./за проект