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php - 返回 MySQL 查询结果时显示 "x results found"或 "no results found"

转载 作者:可可西里 更新时间:2023-11-01 08:48:53 26 4
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搜索功能非常有用,但我不确定如何为其添加某些“附加功能”。当有结果显示时,我还想说:

"Your search returned x results." 

后面是结果。

当没有结果显示时,我想说:

"Your search returned no results.  Please try again."

当用户没有在搜索表单中输入任何内容时,我希望它说:

"You did not enter a search term."

我是一名 PHP 初学者,我不确定如何将其实现到我当前的代码中;我尝试了很多不同的方法,但它要么给我错误,要么在没有结果时返回空白页。

任何指导或帮助都会很棒。谢谢。

这是我当前的代码:

<?php

//STEP 1 Connect To Database
$connect = mysql_connect("localhost","tarb89_admin","leccums");
if (!$connect)
{
die("MySQL could not connect!");
}

$DB = mysql_select_db('tarb89_characters');

if(!$DB)
{
die("MySQL could not select Database!");
}

//STEP 2 Check Valid Information
if(isset($_GET['search']))
{
//STEP 3 Declair Variables
$Search = $_GET['search'];
$result = mysql_query("SELECT * FROM characters WHERE name LIKE '%$Search%' ");

while($row = mysql_fetch_assoc($result))
{
$name = $row['name'];
$id = $row['id'];
$breed = $row['breed'];
$gender = $row['gender'];
$genetics = $row['genetics'];
$profile = $row['profile'];
$player = $row['player'];
$color = $row['color'];
$markings = $row['markings'];
$traits = $row['traits'];
$defects = $row['defects'];
$extras = $row['extras'];

echo " <h3>$name</h3>
<table width='700px' cellpadding='5' cellspacing='0'>
<tr>
<td><p>
<b>ID Number: </b>$id<br>
<b>Breed: </b>$breed<br>
<b>Gender: </b>$gender<br>
<b>Genetics: </b>$genetics<br>
<b>Profile:</b> <a href='$profile'>$profile</a><br>
<b>Player:</b> $player</p></td>
<td><p>
<b>Color:</b> $color<br>
<b>Markings:</b> $markings<br>
<b>Traits:</b> $traits<br>
<b>Defects:</b> $defects<br>
<b>Extras:</b> $extras</p></td>
</table>";

}
}
?>

最佳答案

使用 mysql_num_rows() 返回行数

if($search!="") {
// Your Query

$num = mysql_num_rows($result);
if($num >= 1)
{
echo "Your search returned $num results";
// your code
}
else
{
echo "Your search returned no results. Please try again";
}
} else {
echo "You did not enter a search term";
}


您的代码

if(isset($_GET['search']))
{
//STEP 3 Declair Variables
$Search = $_GET['search'];
$result = mysql_query("SELECT * FROM characters WHERE name LIKE '%$Search%' ");

$num = mysql_num_rows($result);

if($num >= 1)
{
echo "Your search returned $num results";
while($row = mysql_fetch_assoc($result))
{
$name = $row['name'];
$id = $row['id'];
$breed = $row['breed'];
$gender = $row['gender'];
$genetics = $row['genetics'];
$profile = $row['profile'];
$player = $row['player'];
$color = $row['color'];
$markings = $row['markings'];
$traits = $row['traits'];
$defects = $row['defects'];
$extras = $row['extras'];

echo " <h3>$name</h3>
<table width='700px' cellpadding='5' cellspacing='0'>
<tr>
<td><p>
<b>ID Number: </b>$id<br>
<b>Breed: </b>$breed<br>
<b>Gender: </b>$gender<br>
<b>Genetics: </b>$genetics<br>
<b>Profile:</b> <a href='$profile'>$profile</a><br>
<b>Player:</b> $player</p></td>
<td><p>
<b>Color:</b> $color<br>
<b>Markings:</b> $markings<br>
<b>Traits:</b> $traits<br>
<b>Defects:</b> $defects<br>
<b>Extras:</b> $extras</p></td>
</table>";
}
}
else
{
echo "Your search returned no results. Please try again";
}
} else {
echo "You did not enter a search term";
}


注意: mysql_* 函数已弃用,请尽快使用 mysqli_* 函数

关于php - 返回 MySQL 查询结果时显示 "x results found"或 "no results found",我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/19841425/

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