Question

处理项目以显示汽车列表

$username="username";
$password="password";
$database="listofcars";
$mysqli = new mysqli("localhost", $username, $password, $database);
@mysql_select_db($database) or die( "Unable to select database");
$query2="SELECT * FROM cars";
$result=$mysqli->query($query2);
$num=$mysqli->mysqli_num_rows($result);
$mysqli->close();
echo "
<div class="item">
    <div class="container">
        <div class="imgcontainer"><img alt="Cars for sale" src="$carimage" width="380" height="380" /></div>
        <div class="details">
            <a href="$internallink" target="_blank">
                <h3 class="title"> $carname   <br />
                <span> $cartype </span></h3>
                <p>
                    $cardesc
                </p>
                <div class="button"><span data-hover="Order Car">Order Car</span></div>
            </a>
        </div>
    </div>
</div>

我需要它来循环结果

数据库名称：listofcars tablename：cars

字段名称

carsid carname carimage cartype internallink cardesc

Answer 1

您需要保持连接处于打开状态，直到完成后再运行查询。

像

这样的东西

$servername = "localhost";
    $username = "username";
    $password = "password";
    $dbname = "listofcars";

    // Create connection
    $conn = new mysqli($servername, $username, $password, $dbname);

    //Query
    $sql = "SELECT * FROM cars";  

    $result = mysqli_query ($conn, $sql);

     while($row = mysqli_fetch_array($result)) {
     //Whatever you want to do for each row...
     echo "
     <div class='item'>
                    <div class='container'>
                        <div class='imgcontainer'><img alt='Cars for sale' src='".$row[carimage]."' width='380' height='380' /></div>
                        <div class='details'>
                            <a href='".$row[internallink]."' target='_blank'>
                                <h3 class='title'> ".$row[carname]."   <br />
                                <span> ".$row[cartype]." </span></h3>
                                <p>
                                ".$row[cardesc]."
                                </p>
                                <div class='button'><span data-hover='Order Car'>Order Car</span></div>
                            </a>
                        </div>
                    </div>
                </div>"; 


 }

$conn->close();

Answer 2

试试看它是否有效。

<?php
    $servername = "localhost";
    $username = "username";
    $password = "password";
    $dbname = "listofcars";

    // Create connection
    $mysqli = mysqli_connect($servername, $username, $password, $dbname);

    // Check connection
    if (!$mysqli) {
        die("Connection failed: " . mysqli_connect_error());
    }

    $sql = "SELECT * FROM cars";
    $result = mysqli_query($mysqli, $sql);

?>
<?php while($row = mysqli_fetch_assoc($result)) { ?>
<div class="item">
    <div class="container">
        <div class="imgcontainer">
            <img alt="Cars for sale" src="<?php echo $row['carimage']; ?>" width="380" height="380">
            </div>
            <div class="details">
            <a href="<?php echo $row['internallink']; ?>" target="_blank">
                <h3 class="title"><?php echo $row['carname']; ?><br><span><?php echo $row['cartype']; ?></span></h3>
                <p><?php echo $row['cardesc']; ?></p>
                <div class="button">
                    <span data-hover="Order Car">Order Car</span>
                </div>
            </a>
        </div>
    </div>
</div>
<?php } ?>

mysqli_close($mysqli);

显示mysql数据库的结果

2 个答案: