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php - 搜索查询在 PHP 中不起作用

转载 作者:行者123 更新时间:2023-11-29 13:03:58 25 4
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当我使用空字段(字符串)搜索时,我的搜索查询不起作用,它返回所有结果

这是代码:

if(isset($_GET['submit1']))
{
$skills=trim($_GET['skillset']);
$city=trim($_GET['city']);
$state=trim($_GET['state']);
$expinmonth=trim($_GET['expinmonth']);

$result = mysqli_query($dbc,"select * from seeker where (skillset like '%$skills%') or (city like '%$city%') or (state like '%$state%') or (expinmonth like '%$expinmonth%')");

if(!mysqli_num_rows($result) == 0)
{
while($data = mysqli_fetch_array($result))
{
echo .$data['fname'].' '.$data['lname'];
}
}
else
{
echo 'No Results';
}
}

最佳答案

我怀疑您想要做的是在跳过字段时检查空白字段。在这种情况下,像这样:-

<?php

if(isset($_GET['submit1']))
{
$skills = mysqli_real_escape_string($dbc, trim($_GET['skillset']));
$city = mysqli_real_escape_string($dbc, trim($_GET['city']));
$state = mysqli_real_escape_string($dbc, trim($_GET['state']));
$expinmonth = mysqli_real_escape_string($dbc, trim($_GET['expinmonth']));

$result = mysqli_query($dbc,"select *
from seeker
where ".(($skills == '' ) ? "skillset = ''" : "skillset like '%$skills%'")."
or ".(($city == '' ) ? "city = ''" : "city like '%$city%'" )."
or ".(($state == '' ) ? "state = ''" : "state like '%$state%'")."
or ".(($expinmonth == '' ) ? "expinmonth = ''" : "expinmonth like '%$expinmonth%'")." ");

if(!mysqli_num_rows($result) == 0)
{
while($data = mysqli_fetch_array($result))
{
echo .$data['fname'].' '.$data['lname'];
}
}
else
{
echo 'No Results';
}
}

?>

根据 DrCopyPaste 的评论,如果您只想忽略空白输入:-

<?php

if(isset($_GET['submit1']))
{
$skills = mysqli_real_escape_string($dbc, trim($_GET['skillset']));
$city = mysqli_real_escape_string($dbc, trim($_GET['city']));
$state = mysqli_real_escape_string($dbc, trim($_GET['state']));
$expinmonth = mysqli_real_escape_string($dbc, trim($_GET['expinmonth']));

$result = mysqli_query($dbc,"select *
from seeker
where 1=2
".(($skills != '' ) ? " OR skillset like '%$skills%'" : "")."
".(($city != '' ) ? " OR city like '%$city%'" : "")."
".(($state != '' ) ? " OR state like '%$state%'" : "")."
".(($expinmonth != '' ) ? " OR expinmonth like '%$expinmonth%'" : "")." ");

if(!mysqli_num_rows($result) == 0)
{
while($data = mysqli_fetch_array($result))
{
echo .$data['fname'].' '.$data['lname'];
}
}
else
{
echo 'No Results';
}
}

?>

关于php - 搜索查询在 PHP 中不起作用,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/22982266/

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