Question

我试图从mysql数据库中的行创建div，但是当我遍历数据库表并将返回的行应用到div时，我只返回最后一行。这是代码：

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

// Create connection
$conn = new mysqli($servername, $username, $password, $dbname);
// Check connection
if ($conn->connect_error) {
    die("Connection failed: " . $conn->connect_error);
} 

$sql = "SELECT * from getinfo";
$result = $conn->query($sql);

echo " <div id=\"vertical-with-linking\" class=\"fn-container fn-vertical\"         style=\"color: rgb(51, 51, 51); margin: 0px auto; overflow: hidden; position:  relative; width: 580px; height: 280px\">";

if ($result->num_rows > 0) {
    // output data of each row
    while($row = $result->fetch_assoc()) {

        echo "<br> id: ". $row["id"]. " - Name: ". $row["divtitle"]. " " .   $row["datalink"] . "<br>";// works fine

        //echo "<div title = " . $row['divtitle'] . " data-link: " .  $row['datalink'] . "><img src=" . $row['imagesrc'] . " />";//returns last row only
    }
 } else {
    echo "0 results";
 }

 $conn->close();
?>

非常感谢任何建议。

Answer 1

这是一种更安全的方式。您没有关于属性值的引号，如果有$row['divtitle']中有空格，那么它们必须用双引号或单引号引用。

它也应该是data-link=而不是data-link:

你需要在img标签完成后关闭div

echo "<div title='{$row['divtitle']}' data-link='{$row['datalink']}'><img src='{$row['imagesrc']}'/></div>";

Answer 2

你应该关闭div

echo "<div title = " . $row['divtitle'] . " data-link: " .  $row['datalink'] . "><img src=" . $row['imagesrc'] . " />";//returns last row only

到

echo "<div title = " . $row['divtitle'] . " data-link: " .  $row['datalink'] . "><img src=" . $row['imagesrc'] . " /></div>";//returns last row only

Answer 3

我认为问题可能是你的Div没有关闭。

试试这个。

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

// Create connection
$conn = new mysqli($servername, $username, $password, $dbname);
// Check connection
if ($conn->connect_error) {
    die("Connection failed: " . $conn->connect_error);
} 

$sql = "SELECT * from getinfo";
$result = $conn->query($sql);

echo " <div id=\"vertical-with-linking\" class=\"fn-container fn-vertical\"         style=\"color: rgb(51, 51, 51); margin: 0px auto; overflow: hidden; position:  relative; width: 580px; height: 280px\">";

if ($result->num_rows > 0) {
    // output data of each row
    while($row = $result->fetch_assoc()) {

        echo "<br> id: ". $row["id"]. " - Name: ". $row["divtitle"]. " " .   $row["datalink"] . "<br>";// works fine

        echo "<div title = " . $row['divtitle'] . " data-link= " .  $row['datalink'] . "><img src=" . $row['imagesrc'] . " /></div>";//returns last row only
    }

    echo "</div>";
 } else {
    echo "0 results";
 }

 $conn->close();
?>

试图从mysql数据库创建div

3 个答案: