Question

我正在尝试根据数组中包含的标识符从MYSQL数据库中获取记录。要获取标识符，我的代码如下：

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

使用print_r访问数组详细信息时，这似乎工作正常。

我的下一个失败的陈述如下：

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.
    }
}

这只会触发die()语句。我尝试使用implode代替join但没有成功。

编辑：
失败查询后的die($dbc->error)显示错误为：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']);
}

然后

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.
}
}

使用它。我认为它会对你有帮助。

Answer 3

使用implode创建一个字符串并在其周围添加括号。

获取数组中的id值，然后将其内爆。请看下面的示例 - 这可行。

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

Answer 4

编辑：两个查询都可以与JOIN stament结合使用

$query = "
    SELECT
        dd.user_id,
        dd.foo
    FROM
        tmp_agent_coverage as ac
    LEFT JOIN
        tmp_detail_db as dd
    ON
        ac.agent_id=dd.user_id
    WHERE
        ac.primary_area LIKE '$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：自包含示例

<?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);
    }
}

打印

1 fooxy1
3 fooxy2
5 fooxy3

原始答案：
mysqli::query返回false表示执行查询时发生错误。那可能是例如语法错误或权限或...或...
$ dbc对象的error和errno属性应包含有关错误的更详细信息。但是，您不应该向任何用户显示完整的错误消息因此，出于调试目的，定义类似

的函数

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>';
    }
}

然后在您的代码中使用此函数，如

<?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");

在完成调试后删除define(...)行。

Answer 5

你可以这样做

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.
    }
}

希望这有助于

Answer 6

试试这个

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

我在我的localhost中尝试过它，它完美无缺！

MYSQL IN语句语法不起作用

6 个答案: