Question

首先，我很感激my question last week给我的帮助，但我仍然无法理解这一点。

当我在搜索框中输入单词或术语时，由以下代码处理......

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

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

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

$sql = "SELECT ID, FirstName, LastName FROM `table` LIMIT 0, 30 ";
$result = mysqli_query($conn, $sql);

if (mysqli_num_rows($result)>0) { 
  // output data of each row
  while($row=mysqli_fetch_assoc($result)) { 
    echo "ID: " . $row["ID"]. " - Name: " . $row["FirstName"]. " " . $row["LastName"]. "<br>"; 
    } 
} else { 
echo "0 results"; 
} 

mysqli_close($conn);
?>

我从SQL表中列出了所有可能的结果。

那么，如何将结果限制为最终用户输入的结果？我知道我应该使用LIKE或WHERE语句。

如果有帮助，我正在建立一个医疗专业人员数据库，他们的诊所名称和地点。

Answer 1

假设您的表单如下所示：

<form action="" method="post">
    Search: <input name="search" type="text">
</form>

您可以按如下方式扩展SQL：

$sql = "SELECT ID, FirstName, LastName FROM `table` ";
if (isset($_POST['search'])) {
    $search = mysqli_real_escape_string($conn, $_POST['search']);
    $sql .= " WHERE FirstName LIKE '%$search%'
                 OR LastName LIKE '%$search%' ";
}
$sql .= " LIMIT 0, 30 ";
$result = mysqli_query($conn, $sql);
// ... etc.

要清楚，“......等”表示您的输出代码保持原样：

if (mysqli_num_rows($result)>0) { 
    // output data of each row
    while($row=mysqli_fetch_assoc($result)) { 
        echo "ID: " . $row["ID"]. " - Name: " . $row["FirstName"]. " " . $row["LastName"]. "<br>"; 
    } 
} else { 
    echo "0 results"; 
} 

mysqli_close($conn);

Answer 2

$sql1=mysqli_query("SELECT*FROM table ORDER BY id DESC LIMIT 1");
$result1=mysqli_fetch_array($sql1);

从PHP中的SQL表中获取单个结果

2 个答案: