Question

我正在尝试从2个表中检索数据。首先，我从表dealerID中检索了medicinalInfo，然后使用此dealerID我尝试从表users收集此ID的信息，但是，

致命错误：未捕获错误：在C：\ xampp \ htdocs \ Trying \ search.php中调用boolean上的成员函数fetch_assoc（）：16堆栈跟踪：＃0 {main}在C：\ xampp \ htdocs中抛出\在第16行尝试\ search.php

显示此错误..

码

<?php
    $con = mysqli_connect("localhost", "root", "") or  die(mysqli_error($con));
    mysqli_select_db($con,"Ajmal") or die(mysqli_error($con));

    $search = $_POST['search'];

    $sql = "SELECT medicinName,pricerPerSheet,dealerID FROM medicinalinfo WHERE medicinName LIKE '%$search%'";
    $result=$con->query($sql);
    $count=$result->num_rows;
    $val=$result->fetch_array();
    $id=$val['dealerID'];
    //echo $count;

    $sql1 = "SELECT name,email,contact,ststus,distric,place FROM users WHERE id='$id'";
    $query=$con->query($sql1);
    $details=$query->fetch_assoc();
?>

第16行是$details=$query->fetch_assoc();

Answer 1

也许你可以使用这个例子：

<?php
$con = mysqli_connect("localhost", "user", "pass", "database");

/* validate conection */
if (mysqli_connect_errno()) {
    printf("failed: %s\n", mysqli_connect_error());
    exit();
}
$query = "SELECT Name, CountryCode FROM City ORDER by ID DESC LIMIT 50,5";

if ($result = mysqli_query($con, $query)) {

/* get data  */
while ($row = mysqli_fetch_assoc($result)) {
    printf ("%s (%s)\n", $row["Name"], $row["CountryCode"]);
}

/* free resul */
mysqli_free_result($resultado);
}
/* close conection */
mysqli_close($link);
?>

数据表使用Bootstrap，php和ajax

1 个答案: