Php 获取数组到字符串的转换错误

有人能帮我一下这个代码有什么问题吗？我正在尝试将提取的数据存储到一个数组中，我想基于该数组的值进行存储。我收到一个数组到字符串转换的错误。数组的数据类型值为

string

这是密码

            $sql3 ="SELECT DISTINCT subj_descr FROM subj_enrolled WHERE enroll_ref = '$ref'";
           $results = mysqli_query($con, $sql3);
           $data = array();
              while($row = mysqli_fetch_array($results)){
                      $data[] = array($row['subj_descr']);
              }


            $sql ="SELECT * FROM notification WHERE subj_descr IN ({implode(',', $data})";
           $result = mysqli_query($con, $sql);

           $count = mysqli_num_rows($result);

---

删除while循环中的

数组

：

$sql3 ="SELECT DISTINCT subj_descr FROM subj_enrolled WHERE enroll_ref = '$ref'";
       $results = mysqli_query($con, $sql3);
       $data = array();
          while($row = mysqli_fetch_array($results)){
                  $data[] = $row['subj_descr'];
          }


        $sql ="SELECT * FROM notification WHERE subj_descr IN ({implode(',', $data})";
       $result = mysqli_query($con, $sql);

       $count = mysqli_num_rows($result);

---

您正在创建一个

多维

数组，因此请更改此语句

$data[] = array($row['subj_descr']);

到

由于语句中的

SQL始终使用一维数组，所以在查询where子句中也要进行更改

我已全部更改，请尝试以下代码：

 <?php
    $sql3 ="SELECT DISTINCT subj_descr FROM subj_enrolled WHERE enroll_ref = '$ref'";
    $results = mysqli_query($con, $sql3);
    $data = array();
    while($row = mysqli_fetch_array($results)){
        $data[] = $row['subj_descr'];
    }
    $dataStr = implode(',', $data);

    $sql ="SELECT * FROM notification WHERE subj_descr IN (".$dataStr.")";
    $result = mysqli_query($con, $sql);

    $count = mysqli_num_rows($result);
    ?>

创建一个新变量并在其中内爆

试试这个

$implodeAray = implode(",", $data);
$sql ="SELECT * FROM notification WHERE subj_descr IN ($implodeAray)";

循环时，将数组的元素存储在另一个数组中

这样做：

$sql3 ='SELECT DISTINCT subj_descr 
        FROM subj_enrolled 
        WHERE enroll_ref = "$ref"';

$results = mysqli_query($con, $sql3);
$data = array();

while($row = mysqli_fetch_array($results)){
    //Your error was here
    //Each elements is escaped for security reasons
    $data[] = mysqli_escape_string($con,$row['subj_descr']);          
}

//This implodes and puts a single quote around each element
$dataIn= '\'' . implode( '\', \'', $data ) . '\'';

$sql ="SELECT * FROM notification 
       WHERE subj_descr IN ($dataIn)";

$result = mysqli_query($con, $sql);

$count = mysqli_num_rows($result);

subu descr在（{infrade（'，'，$data}）中的位置出错；只需删除while循环中的数组，其余一切都很好，正如@Insomania所建议的那样，只需从这个
数组（$row['subu descr']）
中删除
数组（$row['subu descr']）
；稍微解释一下就可以让他明白了。我得到了一个错误'mysqli num rows（）期望参数1是mysqli_result，布尔值在`中给出，但当我将其更改为` in（'计算机编程1'，'计算机编程2'）；它工作了吗？我试图将
$dataStr
放到
vardump
中，我得到了
字符串（59）“ENGLISH 2，KUMUNIKASYON，计算机编程2，数学2”的输出“
$dataStr有一个值，为什么我得到一个错误？回显此查询并直接在phpmyadmin中运行以检查确切的错误或可能是空的结果表示您的查询是错误的，因此mysqli_num_rows（）期望参数1是mysqli_结果发生错误。或者与数据库中的任何记录都不匹配我得到的错误是`mysqli_num_rows（）期望参数1是mysqli_result，布尔值在`中给出，但当我将其更改为` in（'COMPUTER PROGRAMMING 1'，'COMPUTER PROGRAMMING 2'），它工作了吗？是的，如果数组中有字符串，则需要用引号（'a'，'b'，'c'）分隔它。在`中之前删除`并通过echo$sql显示查询。我得到了一个错误'mysqli_num_rows（）期望参数1是mysqli_result，布尔值在`中给出，但当我将其更改为` in（'计算机编程1'，'计算机编程2'）；它工作了吗？我试图将$dataIn放入vardump，我得到字符串（59）“ENGLISH 2，KUMUNIKASYON，COMPUTER PROGRAMMING 2，MATHEMATICS 2”的输出$dataIn有一个值，为什么我会出错？@wataru我已经更新了代码，将每个元素用一个引号括起来，但是在构建数组之前，必须在每个元素上运行mysql\u escape\u字符串，如
$data[]=mysqli\u escape\u string（$con，$row['subc\u descr']）
；它是空的..但我的数据库有两条记录。您可以再次检查您的查询您可以这样重写您的查询：
$sql=“选择COUNT（*）作为sub descr IN（$dataIn）”通知中的总计；
然后通过这种方式获取结果
$row=mysqli\u fetch assoc（$sql）；echo$row['total']；
$sql3 ='SELECT DISTINCT subj_descr 
        FROM subj_enrolled 
        WHERE enroll_ref = "$ref"';

$results = mysqli_query($con, $sql3);
$data = array();

while($row = mysqli_fetch_array($results)){
    //Your error was here
    //Each elements is escaped for security reasons
    $data[] = mysqli_escape_string($con,$row['subj_descr']);          
}

//This implodes and puts a single quote around each element
$dataIn= '\'' . implode( '\', \'', $data ) . '\'';

$sql ="SELECT * FROM notification 
       WHERE subj_descr IN ($dataIn)";

$result = mysqli_query($con, $sql);

$count = mysqli_num_rows($result);