PHP :Warning: mysqli_num_rows() 期望参数 1 是 mysqli_result，给出的字符串
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PHP :Warning: mysqli_num_rows() expects parameter 1 to be mysqli_result, string given
提问by Sikander
i am trying to fetch data as per following query
我正在尝试按照以下查询获取数据
$query= "SELECT * FROM residential ";
if($type!=""){
$query.=" AND type='$type'";
}
if($unit_type!=""){
$query.=" AND unit_type='$unit_type'";
}
if(($min_price!="") && ($max_price!="")){
$query.=" AND price BETWEEN '$min_price' AND '$max_price' ";
}
if(($min_bedrooms!="") && ($max_bedrooms!="")){
$query.=" AND bedrooms BETWEEN '$min_bedrooms' AND '$max_bedrooms'";
}
if($query==FALSE){
echo mysqli_error($connect);
die;
}
$result= mysqli_query($connect,$query);
this is how i use it
这就是我使用它的方式
<div class="row">
<?php if(mysqli_num_rows($query)>0):?>
<?php while($row= mysqli_fetch_assoc($query)):
print_r($row);
die;
?>
<div class="col-md-4 col-sm-4 col-xs-12">
<div class="row property-listing">
<div class="col-md-6 col-sm-6 col-xs-12">
<img src="images/1.png" class="img-responsive full-width">
</div>
<div class="col-md-6 col-sm-6 col-xs-12 property-desc">
<h3>AED<br/> <?php echo $row['price'];?></h3>
<h5>Unit Type: <?php echo $row['unit_type'];?></h5>
<h5>Available for :<?php echo $row['type'];?></h5>
<h5>Location :<?php echo $row['area'];?></h5>
<h5>Bedrooms :<?php echo $row['bedrooms'];?></h5>
</div>
</div>
</div>
<?php endwhile;?>
<?php else:
echo 'We found no record for your search criteria ';
echo '<a href="index.html">Refine Search</a>'
?>
<?php endif;?>
</div>
This is what i get as error
这是我得到的错误
( ! ) Warning: mysqli_num_rows() expects parameter 1 to be mysqli_result, string given
( ! ) 警告：mysqli_num_rows() 期望参数 1 为 mysqli_result，给出字符串
values being posted and fetched are correct but something is wrong with query , Please help me sort it out
发布和获取的值是正确的，但查询有问题，请帮我解决
Thanks
谢谢
回答by George Pant
$queryis your query string not your result set.So just use $result
$query是您的查询字符串不是您的结果集。所以只需使用 $result
<?php if(mysqli_num_rows($result)>0):?>
<?php while($row= mysqli_fetch_assoc($result)):
回答by Md. Sahadat Hossain
Just change this code
只需更改此代码
$query= "SELECT * FROM residential ";
Into
进入
$query= "SELECT * FROM residential WHERE 1";
And change this code
并更改此代码
<?php if(mysqli_num_rows($query)>0):?>
<?php while($row= mysqli_fetch_assoc($query)):
print_r($row);
die;
?>
To
到
<?php
$res = mysqli_query($con, $query); //replace $con with your db connection variable
if(mysqli_num_rows($res)>0):?>
<?php while($row= mysqli_fetch_assoc($res)):
print_r($row);
die;
?>
回答by Yair.R
You are adding to mysqli_num_rows()and mysqli_fetch_assoc()the query string when it expects mysqli_result- In your case its the variable $result.
要添加到mysqli_num_rows()，并mysqli_fetch_assoc()在预计的查询字符串mysqli_result-在你的情况下，它的变量$result。
回答by Mujahidh
In your Sql query you do not include a WHERE clause.
在您的 Sql 查询中，您不包含 WHERE 子句。
If you want to concatenate your query with other argument for the first sql query add a WHEREclause to your first sql string like this
如果您想将您的查询与第一个 sql 查询的其他参数连接起来，请将WHERE子句添加到您的第一个 sql 字符串中，如下所示
$query= "SELECT * FROM residential WHERE table_id!=0";Here table_idis the primary key of your table.it will be a good practice for concatenating the query.
after that you can add another values to concatenate the query like
$query= "SELECT * FROM residential WHERE table_id!=0";这里table_id是你的表的主键。这将是连接查询的好习惯。之后，您可以添加另一个值来连接查询，如
if($type!=""){
$query.=" AND type='$type'";
}
so your final query something looks like
所以你的最终查询看起来像
SELECT * FROM residential WHERE table_id!=0 AND type='$type'
Next here you are passing $queryto mysqli_num_rowsyou should pass the $resultfor this
接下来，您将传递$query给mysqli_num_rows，您应该$result为此传递
