Php 如果数据库表中有0项，则返回错误

如果表中没有列出任何内容，我正在寻找显示错误消息的方法

我有一张照片桌

如果这个表是空的，我想回显一些东西

否则，请展示图片

在那张桌子里面我有

id, name, url
id = id
name = name of image
url = url of image.

如果没有行，我们有一个错误

$query1 = mysql_query("SELECT COUNT(*) FROM photos;");
mysql_fetch_array($query1);
if(empty($query1)) {
echo "nothing";
} else {
echo "good";
}

试试这个

$query1 = mysql_query("SELECT COUNT(*) FROM photos;");

$result = mysql_fetch_array($query1);

if(empty($result)) {
    echo "nothing";
} else {
    echo "good";
}

试试这个

$query = "SELECT * FROM photos";
$result= mysql_query($query);
$length= mysql_num_rows($result);

if($length>0)
{
   while($rows = mysql_fetch_array($result))
   {
       echo $rows['name'];
       echo "<img src='$rows[url]' />";
   }
}
else
{
   echo "Nothing to display";
}

$query=“从照片中选择*”；
$result=mysql\u query（$query）；
$length=mysql\u num\u行（$result）；
如果（$length>0）
{
while（$rows=mysql\u fetch\u array（$result））
{
echo$rows['name']；
回声“；
}
}
其他的
{
回应“无需显示”；
}

---

希望这能起作用

像

$sql = "SELECT COUNT(*) AS amountPhotos FROM photos";
$result = mysql_query($sql);
$row = mysql_fetch_assoc($result);
if ($row["amountPhotos"] == 0) {
  echo "There are no photos in the photo table.";
}

或

---

这大致概括了这个问题的答案：

他们甚至提供了一个示例代码：

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

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

$sql = "SELECT id, firstname, lastname FROM MyGuests";
$result = $conn->query($sql);

if ($result->num_rows > 0) { //<--- here they check if number of rows returned is greater than 0 (so there is data to display)
    // output data of each row
    while($row = $result->fetch_assoc()) {
        echo "id: " . $row["id"]. " - Name: " . $row["firstname"]. " " . $row["lastname"]. "<br>";
    }
} else {
    echo "0 results"; //<----- nothing found
}
$conn->close();
?> 

---

只要修改这个，你就可以开始了。

谢谢，好问题。我启动了一个查询，从照片中选择计数（*）。如果（query1=0）show error在查询结果上使用

if（）

，

empty（）

，可以添加带问题的代码吗？我选择这个是因为它最简单。其他的很好，但这更容易。谢谢@tobse

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

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

$sql = "SELECT id, firstname, lastname FROM MyGuests";
$result = $conn->query($sql);

if ($result->num_rows > 0) { //<--- here they check if number of rows returned is greater than 0 (so there is data to display)
    // output data of each row
    while($row = $result->fetch_assoc()) {
        echo "id: " . $row["id"]. " - Name: " . $row["firstname"]. " " . $row["lastname"]. "<br>";
    }
} else {
    echo "0 results"; //<----- nothing found
}
$conn->close();
?>