Question

我真的很陌生，正在从事这项作业，我应该在数据库中为作者创建一个下拉列表，并且当用户将鼠标悬停在其中一位作者上时，他们所写的书会显示出来在同一页面上的某个区域中，现在我尝试了所有已知的操作，但是仍然出现此错误。

警告：第21行的C：\ xampp \ htdocs \ demo \ books.php中遇到的非数字值

即使我数据库中的数据全部为文本，也不需要任何数值。

我就是这样提交作业的，但我仍然想知道我做错了什么，我将非常感谢您的帮助

            <?php

              $servername = "localhost";
              $username = "root";
              $password = "";
              $dbname = "books";

                 // variables used in script
              $Author =$_REQUEST["Author"];



                       // Create connection
                        $conn = new mysqli($servername, $username, $password, $dbname);

                       // Check connection
                       if (!$conn) {
                           die("Connection failed: " . mysqli_connect_error());
                       }

                       $sql = "SELECT BookName, BookID FROM `books`"+ 'WHERE AuthorName='+ '"'+'$Author'+'"';
            $sqlData=    mysqli_query($conn, $sql);


                if($sqlData)

                {
                while($row =mysqli_fetch_assoc($sqlData)){
                   echo" <table style='width:100%'>"+ "<tr> <th>BookName</th> <th>BookID</th> <th>AuthorName</th> </tr>"+"<tr> <td>"+ $row[BookName] +
                 "</td><td>"+$row[BookID] +"</td> </tr>";
                }
                echo "</table>" ;

                }





           ?>

这是我的下拉列表所在的整个代码

    <!DOCTYPE html>


                    <html>
                       <head>
                          <meta charset = "utf-8">
                          <title>Books</title>
                          <style >
                             label  { width: 5em; float: left; }
                          </style>



                                <script  >

                                function ShowBook_Info(Author) {
                                    request=new XMLHttpRequest();
                                    request.onreadystatechange=function() {
                                        if (request.readyState==4 && request.status==200) {
                                            document.getElementById("books").innerHTML=request.responseText;
                                        }
                                    }

                                    request.open("GET","books.php?Author="+Author,true);
                                    request.send();
                                }

                            </script>


                       </head>

                           <?php

                          $servername = "localhost";
                          $username = "root";
                          $password = "";
                          $dbname = "books";



                             $sql="SELECT AuthorName FROM `books`";

                                   // Create connection
                                    $conn = new mysqli($servername, $username, $password, $dbname);

                                   // Check connection
                                   if (!$conn) {
                                       die("Connection failed: " . mysqli_connect_error());
                                   }



                               $sqlData=    mysqli_query($conn, $sql);

                               if($sqlData){



                                   echo "<select  onchange='ShowBook_Info(this.value)'>" ;
                                   while($row =mysqli_fetch_assoc($sqlData)){
                                       $Value=$row["AuthorName"];
                                       echo "<option value=";
                                       echo '$value'.'">';
                                       echo $row["AuthorName"];
                                       echo "</option>";


                                   }
                                   echo "</select>" ;


                               } else {
                                   echo "Error: " . $sql . "<br>" . mysqli_error($conn);
                               }



                       ?>


                        <body  onload="ShowBook_Info('Harper Lee');" >



                              <div style="color:blue;" id='books'>
                     This is a test text

                     </div>

                          </form>
                       </body>
                    </html>

Answer 1

您的代码中有很多错误。请检查php的语法，以更好地了解php上的编码

$sql = "SELECT BookName, BookID FROM本书"+ 'WHERE AuthorName='+ '"'+'$Author'+'"';

AuthorName='+ '"'+'$Author'+'"';不正确。只是简单的使用。 `AuthorName ='{$ Author}'“;完整的内容将是：

$sql = "SELECT BookName, BookID FROM书籍`WHERE AuthorName ='{$ Author}'“;

$row[BookName] => $row['BookName']或$row["BookName"]

警告：文本值遇到非数字值

1 个答案: