Javascript 如何从数据库回显If语句
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 此代码Javascript 如何从数据库回显If语句,javascript,php,mysql,if-statement,mysqli,Javascript,Php,Mysql,If Statement,Mysqli,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 此代码 <?php $servername = "localhost"; $username = "root"; $password = ""; $dbname = "databasename"; // Create connection $conn = new mysqli($servername, $username, $password, $dbname); // Check co
<?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();
?>
我认为您需要在if语句中移动echo,以便仅在满足要求时进行echo:
<?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();
?>
我认为您需要在if语句中移动echo,以便仅在满足要求时进行echo:
<?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";
}
$conn->close();
?>
if语句位于echo之前,if语句位于echo之前
if ($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/' . $row['image2']. '" alt="Profile Pic"></a>';
}
}
}
if($result->num\u rows>0)
{
//每行的输出数据
而($row=$result->fetch_assoc())
{
如果(!empty($row['image2']))
{
回声';
}
}
}
if语句在echo之前,if语句在echo之前
if ($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/' . $row['image2']. '" alt="Profile Pic"></a>';
}
}
}
if($result->num\u rows>0)
{
//每行的输出数据
而($row=$result->fetch_assoc())
{
如果(!empty($row['image2']))
{
回声';
}
}
}
谢谢!我还有其他图像在后面有相同的if语句,所以我想重复整个语句。@kenny我认为在这种情况下最好的解决方案是创建一个函数,以避免重复您的代码。@kenny我做了一些补充,这可能会进一步帮助您。谢谢!我还有其他图像在后面有相同的if语句,所以我想重复整个语句。@kenny我认为在这种情况下,最好的解决方案是创建一个函数,以避免重复代码。@kenny我做了一些补充,这可能有助于进一步改进。