在PHP中如何编写代码，以便在条件为真时可以获取所有行数据 [英] In PHP how to write code so that it can fetch all row data if condition is true

问题描述

在php中如何编写代码，以便它可以获取所有行数据，如果条件为真，否则显示未找到消息但只有一次



什么我试过了：

<？php 
 
 $ res = mysqli_query（$ conn， SELECT * FROM email WHERE too ='$ email'）; 
 $ product_count = mysqli_num_rows（$ res）; 
 $ row = mysqli_fetch_array（$ res）; 
 
 
 
 
 if（$ product_count> 0）{
 
 foreach（$ res as $ key => $ row） {
＃code ... 
 if（$ row ['status'] =='删除'）{
＃code ... 
 echo $ row ['message' ]。 
 
} 
其他
 {
 echonot found; 
} 
 
} 
 
 
} 
 
 
？> 

此代码正确显示数据，但也显示其他部分，如果条件不匹配则显示其他部分等于找到的行数。如果rows = 4，那么part运行4次。对不起我的英语不好。请尽快回答我

解决方案

res = mysqli_query（

conn，SELECT * FROM email WHERE too ='

电子邮件'）;

in php how to write code so that it can fetch all row data if condition is true else display not found message but only one time

What I have tried:

<?php

$res=mysqli_query($conn, "SELECT * FROM email WHERE too = '$email'");
$product_count = mysqli_num_rows ($res); 
$row=mysqli_fetch_array($res);

	
	
         
            if ($product_count > 0) {
 
                foreach ($res as $key => $row) {
                    # code...
                    if ($row['status'] == 'Delete') {
                        # code...
                        echo $row['message'];

                    }
                    else
                    {
                        echo "not found";
                    }

                }


	       }
       
    	
?>

this code properly display the data but also display else part and if condition not match then it displays else part equal to number of rows found. if rows = 4 then else part run 4 times. Sorry for my English. Please answer me as soon as possible

解决方案

res=mysqli_query(

conn, "SELECT * FROM email WHERE too = '

email'");

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