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我正在尝试按 ID、名字和姓氏进行表单搜索。我希望用户在搜索字段中输入任何一个并从数据库中获取结果。

这是我使用的实际表格：

<form action="form.php" method="post"> 
<input type="text" name="term" />
<input type="submit" value="Submit" /> 
</form>

这是form.php

<?php
$db_hostname = 'localhost';
$db_username = 'test';
$db_password = 'test';
$db_database = 'test';

// Database Connection String
$con = mysql_connect($db_hostname,$db_username,$db_password);
if (!$con)
  {
  die('Could not connect: ' . mysql_error());
  }

mysql_select_db($db_database, $con);
?>

<!DOCTYPE html>
<html lang="en">
    <head>
        <meta charset="utf-8" />
        <title>SpeedZone Data Search</title>
        <style type="text/css">
table { border-collapse:collapse; }
table td, table th { border:1px solid black;padding:5px; }
tr:nth-child(even) {background: #ffffff}
tr:nth-child(odd) {background: #ff0000}
</style>
    </head>
    <body>
<form action="form.php" method="post">  
<input type="text" name="term" />
<input type="submit" value="Search" />  
</form>  
<?php
if (!empty($_REQUEST['term'])) {

$term = mysql_real_escape_string($_REQUEST['term']);     

$sql = "SELECT * FROM rcrentals WHERE firstname LIKE '%".$term."%' or lastname LIKE '%".$term."%' or id = ".$term;
$r_query = mysql_query($sql); 

echo "<table border='1' cellpadding='5'>";
echo "<tr> <th>ID</th> <th>First Name</th> <th>Last Name</th> <th>Address</th> <th>City</th> <th>State</th> <th>Zip</th> <th>Phone</th> <th>DL</th> <th>Email</th> <th>Car and Controller</th> <th></th> <th></th></tr>";

// loop through results of database query, displaying them in the table
        while ($row = mysql_fetch_array($r_query)){

                // echo out the contents of each row into a table
                echo "<tr>";
                echo '<td>' . $row['id'] . '</td>';
                echo '<td>' . $row['firstname'] . '</td>';
                echo '<td>' . $row['lastname'] . '</td>';
                echo '<td>' . $row['address'] . '</td>';
                echo '<td>' . $row['city'] . '</td>';
                echo '<td>' . $row['st'] . '</td>';
                echo '<td>' . $row['zip'] . '</td>';
                echo '<td>' . $row['phone'] . '</td>';
                echo '<td>' . $row['dl'] . '</td>';
                echo '<td>' . $row['email'] . '</td>';
                echo '<td>' . $row['carcont'] . '</td>';
                echo '<td><a href="edit.php?id=' . $row['id'] . '">Edit</a></td>';
                // echo '<td><a href="delete.php?id=' . $row['id'] . '">Delete</a></td>';
                echo '<td><a href="delete.php?id=' . $row['id'] . '" onclick="return confirm(\'Confirm?\')">Delete</a></td>';
                echo "</tr>"; 
        } 

        // close table>
        echo "</table>"; 

}
?>
    </body>
</html>

我目前有这个：它只是搜索ID。我希望能够输入 ID、名字、姓氏或名字和姓氏（如果可能）。

if (!empty($_REQUEST['term'])) {

$term = mysql_real_escape_string($_REQUEST['term']);     

$sql = "SELECT * FROM rcrentals WHERE firstname LIKE '%".$term."%' or lastname LIKE '%".$term."%' or id = ".$term;

我认为有几件事我需要改变，但我很困惑，迷失在其中，无法解决。请帮忙。

4

2 回答 2

1

$term与 比较时需要加上引号id。否则，如果它不是数字，则会出现语法错误。

$sql = "SELECT * FROM rcrentals WHERE firstname LIKE '%".$term."%' or lastname LIKE '%".$term."%' or id = '".$term."'";

此外，这假设您不使用 0 作为id. 当一个数字与一个字符串进行比较时，该字符串被转换为一个数字，所有非数字字符串都被转换为 0 并且它们将匹配那个id. 如果这是一个问题，你应该先检查是否$term是一个数字。如果不是数字，请使用不包含支票的id查询：

$sql = "SELECT * FROM rcrentals WHERE firstname LIKE '%$term%' or lastname LIKE '%$term%'";
if (is_numeric($term)) {
  $sql .= " or id = $term";
}
于 2013-06-20T01:43:28.717 回答
0

您的查询对我来说看起来很酷。尝试使用大括号分隔 OR 条件

$sql = "SELECT * FROM rcrentals WHERE firstname LIKE '%".$term."%' or (lastname LIKE '%".$term."%') or (id = ".$term);

于 2013-06-20T01:47:48.047 回答