PHP选择具有相同id的不同列

我正在设法弄到所有的议程项目。首先，我有一个起始日期为$start的输入，但是当我有输出post_id 6和14 post_id时，我想得到另一个具有相同post_id的meta_值

我想在这里拿到邮政编码6。

如何获取Json中具有相同post\u id的meta\u值？

$sql=从wp\u postemta中选择post\u id，其中meta\u key='eap\u FROM\u day'和'$start'

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这就是您想要的？

我必须在while语句中执行一个新的查询：

$start = $_GET["startDate"]." "."00:00:00";
$end = $_GET["endDate"]." "."00:00:00";
/* Template Name: Example Template */
//            Events as Posts plugin
$servername = "localhost";
$username = "root";
$password = "";
$dbname = "test";
// conectie word geregeld met de database
$conn = mysqli_connect($servername, $username, $password, $dbname);
$sql = "SELECT post_id FROM wp_postmeta WHERE meta_key = '_eventorganiser_schedule_start_start' AND '$start' <= DATE(meta_value)AND '$end' >= DATE(meta_value)";
if ($conn->connect_error) {
    die("Connection failed: " . $conn->connect_error);
}

$all = mysqli_query($conn, $sql);
while($row = mysqli_fetch_assoc($all)) {
    foreach ($row as $r) {
        $sql2 = "SELECT * FROM wp_postmeta WHERE post_id = '$r'";
        $everything = mysqli_query($conn, $sql2);
        while($col = mysqli_fetch_assoc($everything)) {
            $sql3 = "SELECT post_title, post_content FROM wp_posts WHERE ID = '$r'";
            $last = mysqli_query($conn, $sql3);
            while ($rest = mysqli_fetch_assoc($last)) {
                $title= $rest["post_title"];
                $info= $rest["post_content"];

                if ($col["meta_key"] == "_eventorganiser_schedule_start_start") {
                    $date = $col["meta_value"];
                }
                if ($col["meta_key"] == "_eventorganiser_schedule_last_finish") {
                 $enddate = $col["meta_value"];
                }
                $result = array("title"=>$title,"info"=>$info,"begindatum"=> $date,"einddatum"=>$enddate);
            }
        }print_r(json_encode($result));
    }
}

---

将其添加到WHERE子句中为类似的内容编写您自己的查询不是一个非常好的主意，您应该使用WP为此提供的功能@JayBlanchard如果我这样做了，我只会选择post_id你说我想得到其他Collumn的确切意思是什么？是否只需要其他列的meta_键名称？请更新您的问题，以包括所需的代码结果/输出，以及将post_id设置为6的位置。它已经在选择post_id 6。我只想选择其他具有post_id 6AAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA*

$sql = "SELECT * FROM wp_postmeta WHERE meta_key='eap_from_day' AND '$start' <= DATE(meta_value) AND post_id='6' ";

$start = $_GET["startDate"]." "."00:00:00";
$end = $_GET["endDate"]." "."00:00:00";
/* Template Name: Example Template */
//            Events as Posts plugin
$servername = "localhost";
$username = "root";
$password = "";
$dbname = "test";
// conectie word geregeld met de database
$conn = mysqli_connect($servername, $username, $password, $dbname);
$sql = "SELECT post_id FROM wp_postmeta WHERE meta_key = '_eventorganiser_schedule_start_start' AND '$start' <= DATE(meta_value)AND '$end' >= DATE(meta_value)";
if ($conn->connect_error) {
    die("Connection failed: " . $conn->connect_error);
}

$all = mysqli_query($conn, $sql);
while($row = mysqli_fetch_assoc($all)) {
    foreach ($row as $r) {
        $sql2 = "SELECT * FROM wp_postmeta WHERE post_id = '$r'";
        $everything = mysqli_query($conn, $sql2);
        while($col = mysqli_fetch_assoc($everything)) {
            $sql3 = "SELECT post_title, post_content FROM wp_posts WHERE ID = '$r'";
            $last = mysqli_query($conn, $sql3);
            while ($rest = mysqli_fetch_assoc($last)) {
                $title= $rest["post_title"];
                $info= $rest["post_content"];

                if ($col["meta_key"] == "_eventorganiser_schedule_start_start") {
                    $date = $col["meta_value"];
                }
                if ($col["meta_key"] == "_eventorganiser_schedule_last_finish") {
                 $enddate = $col["meta_value"];
                }
                $result = array("title"=>$title,"info"=>$info,"begindatum"=> $date,"einddatum"=>$enddate);
            }
        }print_r(json_encode($result));
    }
}