Question

当我按下我的html页面上的搜索按钮时，它会在我的桌子上添加一辆新车，但我希望能够从我在MySql上创建的汽车列表中搜索汽车。我有一个car_displayed代码，一旦搜索到位，它将在一个表格中显示汽车。

<!DOCTYPE html>
<html lang="en">
<head>
<meta charset="utf-8"/>
<meta name="description" content="Searing for a car" />
<meta name="keywords" content="PHP, MySql" />
<title>Retrieving records to HTML</title>
</head>
<body>
    <h1>Car Search</h1>
<?php
    require_once ("settings.php"); //connection info
    $conn = @mysqli_connect($host,
        $user,
        $pwd,
        $sql_db
    );
    $make = trim($_POST["carmake"]);
    $model = trim($_POST["carmodel"]);
    $price = trim($_POST["price"]);
    $yom = trim($_POST["yom"]);
    $sql_table="cars";
    $query = "insert into $sql_table (make, model, price) values ('$make', '$model', '$price')";
    //execute the query -we should really check to see if the batabase exists first.
    $result = mysqli_query($conn, $query);
    //checks if the exeution was succcessful
    if (!$result) {
            echo "<p class=\"wrong\">Something is wrong with ", $query, "</p>";
            // would not show in a production script
        } else {
            // display an operation successful message
            echo "<p class=\"ok\">Successfully Searched Car record</p>";
        } // if successful query operation

        // close the database connection
        mysqli_close($conn);
     // if successful database connection
?>
</body>
</html>

Answer 1

不是插入新车，而是使用像

这样的'select`查询来查找

$query = "select model, price from $sql_table where make LIKE %$make%"

执行后，$result将包含所请求的信息。

PS：我猜测make是你桌上的主键。

在php中搜索项目

1 个答案: