Синтаксис синтаксиса MYSQL IN не работает

Вы можете сделать это так

if(!empty($agent_primary))
{
    $ids=implode($agent_primary,",");
    $query5="SELECT * FROM detail_db WHERE user_id IN ($ids)";
    $result5=mysqli_query($dbc, $query5) or die("Could not do SELECT");

    while($row=mysqli_fetch_array($result5))
    {
    //do stuff with results here.
    }
}

Надеюсь это поможет

$query="SELECT agent_id FROM agent_coverage WHERE primary_area LIKE '$search_term%'";
$result=mysqli_query($dbc, $query) or die("Could not obtain primary area coverage2");

$agent_primary=array();    

while($r=mysqli_fetch_array($result))
{
  array_push($agent_primary,$r['agent_id']);
}

тогда

if(!empty($agent_primary))
{
$ids=$agent_primary;
$query5="SELECT * FROM detail_db WHERE user_id IN($ids)";
$result5=mysqli_query($dbc, $query5) or die("Could not do SELECT");

while($row=mysqli_fetch_array($result5))
{
//do stuff with results here.
}
}

использовать это. Я думаю, это поможет вам.

edit: Оба запроса могут быть объединены с строкой JOIN

$query = "
    SELECT
        dd.user_id,
        dd.foo
    FROM
        tmp_agent_coverage as ac
    LEFT JOIN
        tmp_detail_db as dd
    ON
        ac.agent_id=dd.user_id
    WHERE
        ac.primary_area LIKE '$search_term%'
";
$result=mysqli_query($dbc, $query) or mysqliError($dbc, "Could not obtain primary area coverage2");
while($r=mysqli_fetch_array($result))
{
    echo $r['user_id'], ' ', $r['foo'], "\n";
}

edit2: автономный пример

<?php
$dbc = new mysqli('localhost', 'localonly', 'localonly', 'test');
if ($dbc->connect_error) {
    var_dump($mysqli->connect_errno, $mysqli->connect_error);
    die;
}
setup($dbc);
$search_term = 'xy';

$query = "
    SELECT
        dd.user_id,
        dd.foo
    FROM
        tmp_agent_coverage as ac
    LEFT JOIN
        tmp_detail_db as dd
    ON
        ac.agent_id=dd.user_id
    WHERE
        ac.primary_area LIKE '$search_term%'
";
$result=mysqli_query($dbc, $query) or mysqliError($dbc, "Could not obtain primary area coverage2");
while($r=mysqli_fetch_array($result))
{
    echo $r['user_id'], ' ', $r['foo'], "\n";
}

define('DEVELOPMENT_DEBUG_MESSAGES', 1);
function mysqliError($dbc, $description) {
    if ( !defined('DEVELOPMENT_DEBUG_MESSAGES') || !DEVELOPMENT_DEBUG_MESSAGES ) {
        echo '<div class="error">', htmlspecialchars($description), '</div>';
    }
    else {
        echo '
', htmlspecialchars($description), '
',
            htmlspecialchars($dbc->error), '</div>';
    }
}

function setup($dbc) {
    $q = array(
        'CREATE TEMPORARY TABLE tmp_agent_coverage (
            agent_id int auto_increment,
            primary_area varchar(32),
            primary key(agent_id),
            key(primary_area)
        )',
        "INSERT INTO tmp_agent_coverage (primary_area) VALUES ('xy1'),('dfg'),('xy2'),('abc'),('xy3')",
        'CREATE TEMPORARY TABLE tmp_detail_db (
            user_id int auto_increment,
            foo varchar(32),
            primary key(user_id)
        )',
        "INSERT INTO tmp_detail_db (foo) VALUES ('fooxy1'),('foodfg'),('fooxy2'),('fooabc'),('fooxy3')"
    );
    foreach($q as $query) {
        $dbc->query($query) or die(__LINE__ .' '.$query. ' '. $dbc->error);
    }
}

prints

1 fooxy1
3 fooxy2
5 fooxy3

original answer:
mysqli::query returning false means that an error occured while executing the query. That could be e.g. a syntax error or privileges or ... or ...
The error and errno properties of your $dbc object should hold more detailed information about the error. But you shouldn't show the complete error message to just any arbitrary user.
So, for debugging purposes define a function like

define('DEVELOPMENT_DEBUG_MESSAGES', 1);
function mysqliError($dbc, $description) {
    if ( !defined('DEVELOPMENT_DEBUG_MESSAGES') || !DEVELOPMENT_DEBUG_MESSAGES ) {
        echo '<div class="error">', htmlspecialchars($description), '</div>';
    }
    else {
        echo '
', htmlspecialchars($description), '
',
            htmlspecialchars($dbc->error), '</div>';
    }
}

and then use this function in your code like

<?php
define('DEVELOPMENT_DEBUG_MESSAGES', 1);
[...]
$query="SELECT agent_id FROM agent_coverage WHERE primary_area LIKE '$search_term%'";
$result=mysqli_query($dbc, $query) or mysqliError($dbc, "Could not obtain primary area coverage2");

remove the define(...) line when you're done debugging.

попробуй это

 $ids=join("','", $agent_primary);
 $query5="SELECT * FROM detail_db WHERE user_id IN('$ids')";

Я попробовал его в своем локальном хосте, и он отлично работает!

Используйте implode для создания строки и добавления скобок вокруг нее.

Получите значения идентификаторов в массиве, а затем взорвите его. Посмотрите ниже пример - это работает.

 $arr = array(1,297,298);
    $new = '(';
    $new .= implode(',',$arr);
    $new .= ')';
    $query = "Select * from wp_posts where ID IN $new";