MYSQL IN语句语法不起作用

Question

I am trying to obtain records from a MYSQL database based on identifiers contained in an array. To get the identifiers my code is as follows:

$query="SELECT agent_id FROM agent_coverage WHERE primary_area LIKE '$search_term%'";
$result=mysqli_query($dbc, $query) or die("Could not obtain primary area coverage2");
while($r=mysqli_fetch_array($result))
{
    $agent_primary[]=$r['agent_id'];
}

This seems to work fine when using print_r to access the array details.

My next statement which is the one that fails is as follows:

if(!empty($agent_primary))
{
    $ids=join(",", $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.
    }
}

This just fires the die() statement. I have tried using implode instead of join with no success.

edit:
die($dbc->error) after the failing query reveals the error to be: unknown column '' in where clause

Answer 1

尝试：

$agent_primary[]="'".$dbc->real_escape_string($r['agent_id'])."'";

Answer 2

$query="SELECT agent_id FROM agent_coverage WHERE primary_area LIKE '$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']);
}

then

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.
}
}

use this. I think it will help you.

Answer 3

Use implode to create a string and append brackets around it.

Get values of ids in a array and then implode it. Just see below example - This works.

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

Answer 4

edit: Both queries can be combined with a JOIN stament

$query = "
    SELECT
        dd.user_id,
        dd.foo
    FROM
        tmp_agent_coverage as ac
    LEFT JOIN
        tmp_detail_db as dd
    ON
        ac.agent_id=dd.user_id
    WHERE
        ac.primary_area LIKE '$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: self-contained example

<?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 '<fieldset class="error"><legend>', htmlspecialchars($description), '</legend>',
            htmlspecialchars($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 eg 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 '<fieldset class="error"><legend>', htmlspecialchars($description), '</legend>',
            htmlspecialchars($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.

Answer 5

You can do it like this

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.
    }
}

Hope this Helps

Answer 6

try this

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

i tried it in my localhost and it works perfect!