javascript - 如何从数据库中回显 If 语句

Question

我在 while 循环内的 IF 语句中显示信息时遇到问题。甚至可以在 while 循环中回显 if 语句吗？请帮忙!

这段代码

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

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

$searchEscaped = $conn->real_escape_string($_GET['username']);
if ($conn->connect_error) {
     die("Connection failed: " . $conn->connect_error);
} 

$sql = "SELECT * FROM users WHERE username = '$searchEscaped' ";
$result = $conn->query($sql);

if ($result->num_rows > 0) {
     // output data of each row
     while($row = $result->fetch_assoc()) {
      echo "
    **if(!empty($row['image2'])) { 
    <a class='example-image-link' href='pictures/".$row['image2']."' data-lightbox='example-set'><img class='example-image'src='pictures/".$row['image2']."'  alt='Profile Pic'></a>
     }
      ";}
} else {
     echo "No users found";
}

$conn->close();
?>  

Answer 1

我认为您需要将 echo 移到 if 语句中，以便仅在满足要求时才回显:

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

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

$searchEscaped = $conn->real_escape_string($_GET['username']);
if ($conn->connect_error) {
    die("Connection failed: " . $conn->connect_error);
}

$sql = "SELECT * FROM users WHERE username = '$searchEscaped' ";
$result = $conn->query($sql);

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

        if (!empty($row['image2'])) {
            echo "
    <a class='example-image-link' href='pictures/" . $row['image2'] . "' data-lightbox='example-set'><img class='example-image'src='pictures/" . $row['image2'] . "'  alt='Profile Pic'></a>

      ";
        }
    }
} else {
    echo "No users found";
}

$conn->close();
?>  

这是一个可以在整个脚本中使用的函数示例:

    function print_image($row_image){
        if(!empty($row_image)){
            return  "
        <a class='example-image-link' href='pictures/" . $row_image . "' data-lightbox='example-set'><img class='example-image'src='pictures/" . $row_image . "'  alt='Profile Pic'></a>

          ";
        }
else{
        return "";
    }
    }

您可以在脚本的任何其他地方调用此函数并执行以下操作:

echo print_image($row['image2']);

或

echo print_image($row['image3']);