我遇到了一些问题，希望你的家伙可以帮助我。
我有这个代码，我无法开始工作。
看，奇怪的是，如果我改变了
$query2 = mysqli_query($dblogin,
"SELECT * FROM devices
RIGHT JOIN repair ON devices.name = repair.modelid
WHERE repair.modelid=$check");
到
$query2 = mysqli_query($dblogin,
"SELECT * FROM devices
RIGHT JOIN repair ON devices.name = repair.modelid
WHERE repair.modelid LIKE %iPad%");
我会工作，很好： - /
我真的输了，我试着回应$ check，看看$ check变量是否有错误。但这样做很好。
<div id="slidingDiv<?=$row["id"] ?>" class="toggleDiv row-fluid single-project">
<div class="span6">
<img src="adm/images/<?=$row["image"] ?>" alt="project <?=$row["id"] ?>" />
</div>
<div class="span6">
<div class="project-description">
<div class="project-title clearfix">
<h3><?=$row["name"] ?></h3>
<span class="show_hide close">
<i class="icon-cancel"></i>
</span>
</div>
<div class="project-info">
<?php
$check = $row["name"];
$query2 = mysqli_query($dblogin,
"SELECT * FROM devices
RIGHT JOIN repair ON devices.name = repair.modelid
WHERE repair.modelid=$check");
while($row = mysqli_fetch_assoc($query2))
{
?>
<div>
<span><?=$row["price"] ?> ,-</span><?=$row["name"] ?>
</div>
<?php
}
?>
<div>
<span>Client</span>Some Client Name
</div>
<div>
<span>Date</span>July 2013
</div>
<div>
<span>Skills</span>HTML5, CSS3, JavaScript
</div>
<div>
<span>Link</span>http://examplecomp.com
</div>
</div>
<p>Believe in yourself! Have faith in your abilities! Without a humble but reasonable confidence in your own powers you cannot be successful or happy.</p>
</div>
</div>
</div>
如果我更改了它，它运行得很好
<?
$query = mysqli_query($dblogin, "select * from brands ORDER BY id");
while($row = mysqli_fetch_assoc($query))
{
?>
<li class="filter" data-filter="<?=$row["name"] ?>">
<a href="#noAction"><img class="aimg" style="height:50px;" src="adm/images/<?=$row["image"] ?>" /></a>
</li>
<?
}
?>
</ul>
<!-- Start details for portfolio project 1 -->
<div id="single-project">
<?
$query = mysqli_query($dblogin, "select * from devices ORDER BY name DESC");
while($row = mysqli_fetch_assoc($query))
{
?>
<div id="slidingDiv<?=$row["id"] ?>" class="toggleDiv row-fluid single-project">
<div class="span6">
<img src="adm/images/<?=$row["image"] ?>" alt="project <?=$row["id"] ?>" />
</div>
<div class="span6">
<div class="project-description">
<div class="project-title clearfix">
<h3><?=$row["name"] ?></h3>
<span class="show_hide close">
<i class="icon-cancel"></i>
</span>
</div>
<div class="project-info">
<?php
$check = $row["name"];
$query2 = mysqli_query($dblogin, "select * from devices right join repair on devices.name = repair.modelid WHERE repair.modelid LIKE '%iPhone 6%'");
while($row = mysqli_fetch_assoc($query2))
{
?>
<div>
<span><?=$row["price"] ?> ,-</span><?=$row["name"] ?>
</div>
<?php
}
?>
</div>
</div>
</div>
</div>
<?
}
?>
修改
SELECT * FROM devices
RIGHT JOIN repair ON devices.name = repair.modelid
WHERE repair.modelid LIKE '%iPhone 6%'
LIMIT 0 , 30
给出这个结果
id name brand image model-id id name modelid price brand date
7 iPhone 6 Apple iphone6_small.jpg 217 Front Glas Udskiftning iPhone 6 1300 1432215986
7 iPhone 6 Apple iphone6_small.jpg 218 Bagside (komplet) iPhone 6 2500 1432216016
7 iPhone 6 Apple iphone6_small.jpg 219 Tænd/sluk Udskiftning iPhone 6 500 1432216041
7 iPhone 6 Apple iphone6_small.jpg 220 Homeknap iPhone 6 500 1432216064
7 iPhone 6 Apple iphone6_small.jpg 221 Ørehøjtaler iPhone 6 500 1432216085
7 iPhone 6 Apple iphone6_small.jpg 222 Ladestik iPhone 6 500 1432216107
7 iPhone 6 Apple iphone6_small.jpg 223 Batteri iPhone 6 500 1432216124
7 iPhone 6 Apple iphone6_small.jpg 224 Vibrator iPhone 6 500 1432216136
7 iPhone 6 Apple iphone6_small.jpg 225 Mikrofon iPhone 6 500 1432216165
7 iPhone 6 Apple iphone6_small.jpg 226 Kamera iPhone 6 500 1432216177
7 iPhone 6 Apple iphone6_small.jpg 227 Højtaler (musik/lyd) iPhone 6 500 1432216191
7 iPhone 6 Apple iphone6_small.jpg 228 WIFI Antenne iPhone 6 500 1432218537
7 iPhone 6 Apple iphone6_small.jpg 229 Jackstik iPhone 6 500 1432218564
7 iPhone 6 Apple iphone6_small.jpg 230 Jailbreak iPhone 6 500 1432218593
7 iPhone 6 Apple iphone6_small.jpg 231 Bagside (komplet) iPhone 6 2500 1432218612
答案 0 :(得分：1)
错别字：
... WHERE repair.modelid LIKE %iPad%"
您没有引用LIKE条款：
... WHERE repair.modelid LIKE '%iPad%'"
^------^--
没有引号意味着%iPad%将被视为未知/非法字段名称，从而导致SQL语法错误。由于您的查询绝对没有错误处理，因此您只是盲目地假设没有错。从不假设成功。总是假设失败，检查失败，并将成功视为一个惊喜。
即使是像这样的一些超级骨干错误处理也会有所帮助：
$result = mysql_query(...) or die(mysql_error());
^^^^^^^^^^^^^^^^^^^^^^
答案 1 :(得分：0)
请原谅我，如果我的PHP有点生疏，但不应该这样：
$query2 = mysqli_query($dblogin, "select * from devices right join repair on devices.name = repair.modelid WHERE repair.modelid='" . $check . "'");
因此，在$ check为“bucket”的示例中，这将运行SQL：
select * from devices right join repair on devices.name = repair.modelid WHERE repaid.modelid='bucket'
答案 2 :(得分：0)
如果$ check仅包含产品名称，例如“iPad”，请更改您的查询：
$query2 = mysqli_query($dblogin, "select * from devices right join repair on devices.name = repair.modelid WHERE repair.modelid='$check'");
更新：
或者LIKE：
$query2 = mysqli_query($dblogin, "select * from devices right join repair on devices.name = repair.modelid WHERE repair.modelid LIKE '%$check%'");
请注意，如果$ check来自用户，则此语句易受SQL注入攻击。
答案 3 :(得分：0)
试试这个。括号周围的括号。虽然你把它称为id，但它等于name，所以我假设不是数字。
$query2 = mysqli_query($dblogin,
"SELECT * FROM devices
RIGHT JOIN repair ON devices.name = repair.modelid
WHERE repair.modelid='{$check}'");