SQL函数没有返回结果[重复]

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mysql_fetch_array()/mysql_fetch_assoc()/mysql_fetch_row()/mysql_num_rows etc… expects parameter 1 to be resource 31 answers
mysql_fetch_array（）/ mysql_fetch_assoc（）/ mysql_fetch_row（）/ mysql_num_rows等...期望参数1为资源31答案

I have a function that takes an SQL query and parameters/array of parameters. This function working fine on UPDATE statements but not on SELECT statements.The function mysqli_fetch_array is not returning anything and I cannot figure out why. The error message I am getting is;

我有一个函数，它接受SQL查询和参数/参数数组。这个函数在UPDATE语句上工作正常但在SELECT语句上没有。函数mysqli_fetch_array没有返回任何内容，我无法弄清楚原因。我得到的错误信息是;

"mysqli_fetch_array() expects parameter 1 to be mysqli_result, boolean given"

“mysqli_fetch_array（）期望参数1为mysqli_result，boolean给出”

I am using PHP version 5.5.

我使用的是PHP 5.5版。

The query is as follows:

查询如下：

<?php

        $txtSearch = isset($_POST["txtSearch"]) ? $_POST["txtSearch"] : "";

            $result_search = null;
            $query = "";
            if(!empty($txtSearch))
            {
                if($page == "adverts")
                {
                    $query = "
                        SELECT
                            A.ID,
                            A.title,
                            A.productname,
                            A.price,
                            A.description,
                            A.productlocation,
                            (SELECT I.name FROM image I WHERE I.advertID = A.ID LIMIT 1) AS name
                        FROM advert A
                        WHERE A.title LIKE ?";
                }
                $result_search = queryDB($query, array("s", "%".$txtSearch."%"));
            }
           while($row = mysqli_fetch_array($result_search))
                {

                                                echo 
                                                    "<div class='post_container'>
                                                            <div class='title_banner'>".$row["title"]."</div>
                                                            <div>£".$row["price"]."</div>
                                                    </div>";
                                            }                           
        ?>

2 个解决方案

#1

You could return $stmt->get_result()instead, which will provide you with a result you can iterate over.

您可以返回$ stmt-> get_result（），它将为您提供可以迭代的结果。

Your error tells you that you are returning a boolean, not a mysqli_result. The documentation on mysqli_stmt::fetch() confirms that it returns a boolean.

您的错误告诉您正在返回布尔值，而不是mysqli_result。 mysqli_stmt :: fetch（）上的文档确认它返回一个布尔值。

#2

Try to check query result before you're passing it as an argument. If it returns false, than there is something wrong with the query.

在将查询结果作为参数传递之前，请尝试检查查询结果。如果返回false，则查询出错。

if($result_search === FALSE) { 
    die(mysql_error()); // TODO: better error handling
}

And I would change your query on something like this:

我会改变你的查询，如下所示：

SELECT
    A.ID, A.title, A.productname,
    A.price, A.description,
    A.productlocation,
    I.name
FROM 
    advert A
INNER JOIN
    image I ON I.advertID = A.ID
WHERE 
    A.title LIKE ?"

#1