Question

我有一个PHP搜索。目前，当没有显示结果时，脚本显示和错误。如何让它显示一条消息，例如＆＃34;未找到结果＆＃34;什么都没有回来？

<?php
include("incls/connector.php"); 

$result = mysql_query("SELECT * FROM user WHERE username='$_POST[compName]'");

while($rowval = mysql_fetch_array($result))
    {
        $complainer = $rowval['username'];
        $department = $rowval['department'];
        $phonenumber = $rowval['phone_number'];
    }
        else
    {
        echo 'No Results were found';
    }

    mysql_close()
?>

Answer 1

几点建议：

停止使用mysql_ *扩展，在PHP 7中弃用并关闭，使用mysqli_*或PDO。
如果需要，尝试调试print_r($_POST)。
您必须使用IF作为ELSE条件。
当您在查询语句中直接使用HTML表单输入时，您需要学习mysqli::real_escape_string

以下是使用 MYSQLi面向对象的代码的完整示例：

示例：

<?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); } $yourInput = $conn->real_escape_string($_POST['compName']); $sql = "SELECT * FROM user WHERE username='$yourInput'"; $result = $conn->query($sql); if ($result->num_rows > 0) { // output data of each row while($row = $result->fetch_assoc()) { //your stuff } } else { echo "no record found"; } $conn->close(); ?>

Answer 2

请试试这个

include("incls/connector.php"); 

$result = mysql_query("SELECT * FROM user WHERE username='$_POST[compName]'");

$num_rows = mysql_num_rows($result); //Check for total number of rows

if($num_rows){
 while($rowval = mysql_fetch_array($result))
    {
        $complainer = $rowval['username'];
        $department = $rowval['department'];
        $phonenumber = $rowval['phone_number'];
    }   
  }
else { echo 'No Results were found'; }

mysql_close();

Answer 3

警告停止使用mysql_ *使用mysqli_ *或PDO
您的代码有误，需要if才能使用else

include("incls/connector.php"); 

$result = mysql_query("SELECT * FROM user WHERE username='".$_POST['compName']."'");//quote for post array

$num_rows = mysql_num_rows($result); //Check for total number of rows

if(is_int($num_rows) && $num_rows >0){
 while($rowval = mysql_fetch_array($result))
    {
        $complainer = $rowval['username'];
        $department = $rowval['department'];
        $phonenumber = $rowval['phone_number'];
    }   
  }
else { echo 'No Results were found'; }

mysql_close();

Answer 4

代码中存在语法错误。 Else没有if

if(count($result)==0){
   echo 'No Results were found';
}

此外，checkout sql injection你传入的行参数可以导致sql注入

Answer 5

请使用if for else statement

if( $rowval ) {
} else {
   echo "No Rows Found";
}

Answer 6

首先获取记录数以显示记录。还要在try catch块中添加代码片段以跟踪异常，而不在UI上显示它们。

try {
$result = mysql_query("SELECT * FROM user WHERE username='$_POST[compName]'");

$count = mysql_num_rows($result); //get the count of rows
if($count > 0)
  {
        while($rowval = mysql_fetch_array($result))
       {
        $complainer = $rowval['username'];
        $department = $rowval['department'];
        $phonenumber = $rowval['phone_number'];
       }
   }    
   else
   {
        echo 'No Results were found';
   }    
    mysql_close();  // Missing semicolon in your code
} catch (e){
   print_r("Error while getting the results.");
}
?>

验证显示消息＆＃34;未找到结果＆＃34;

6 个答案: