Php 意外'；a'；在postman中运行时的json o/p

像这样试试，改变你的方法，发帖子，在身体里传递电子邮件

<?php
    $connect = mysqli_connect("localhost","root","","test");

    echo "aa";

    if(isset($_GET['emailid'])) {
        echo "aa";
        $emailid = $_GET['emailid'];
        $sql  = "SELECT emailid, phoneno, fname FROM registered WHERE emailid = '$emailid'";

        $query = mysqli_query($connect,$sql);
        $resultfetch = mysqli_fetch_array($query);
        $result = array();

        array_push($result, array(
                              "Emailid" => $resultfetch['emailid'],
                              "Phoneno" => $resultfetch['phoneno'],
                              "first name" => $resultfetch['fname']
        ));

        echo json_encode(array("result"=>$result));
        echo json_last_error_msg();
        mysqli_close($connect);
    }

?>

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意外的“a”或“u”？发布响应它显示意外的“a”，但当我将光标放在十字标记上时，它会弹出意外的“u”您正在使用的url？很可能是您的表结构有问题，您确定表的结构良好吗？发布该Web服务api的api（url链接）您如何在标题中传递emailid？get method的Body选项已禁用是您应该在左侧下拉列表中更改post method。。。如果您想在Get方法中传递值，则意味着您应该只在Url中传递值，而不是在标题中传递值。抱歉，您没有完整阅读您的代码，您将Get替换为post，并且在更改后，会提供所需的o/p thanx budyyy:）但我想从用户处获取数据，以便使用post是合适的？？您得到结果了吗？

if(isset($_POST['emailid'])) {
        echo "aa";
        $emailid = $_POST['emailid'];
        $sql  = "SELECT emailid, phoneno, fname FROM registered WHERE emailid = '$emailid'";

        $query = mysqli_query($connect,$sql);
        $resultfetch = mysqli_fetch_array($query);
        $result = array();

        array_push($result, array(
                              "Emailid" => $resultfetch['emailid'],
                              "Phoneno" => $resultfetch['phoneno'],
                              "first name" => $resultfetch['fname']
        ));

        echo json_encode(array("result"=>$result));
        echo json_last_error_msg();
        mysqli_close($connect);
    }