Question

我继续得到这个错误=警告：mysql_fetch_array（）期望参数1是资源，在此代码中给出布尔值：`

 <script>
        //this will trigger automatically when they change the first select box
        $('#dateFilter').on('change', function(event){
        if($(this).val() == 'dateFilter'){
            $("#ajax_reply_div").empty()
        }else{
            var values = $(this).serialize();
            $.ajax({
                url: "AdvanceOption.php",
                type: "post",
                data: values,
                success: function(data){
                $("#ajax_reply_div").empty().append(data);
            },
            error:function(){
                $("#ajax_reply_div").empty().append('something went wrong');
            }
        });
    }
});
</script>

    <div id='center'>
        <?php include  "header.php"; // display menu options ?>



                <form id = "project_form" method="post">
                    <div id="searchPanel">
                        <?php 
                            echo "Choose a date"; 
                            $something_date = $sample->retrieveFilterDate(); 
                            $data_date = array();
                            echo  "<select id='dateFilter' class='dateFilter' name='dateFilter'>";
                            while($row_date = mysql_fetch_assoc($something_date)){
                              array_push($data_date, $row_date);
                              echo  "<option value='".$row_date['date']."' >".$row_date['date']." </option>";
                              } 
                            echo "</select>";

                            echo "\tChoose a place";
                            $something_place = $rm->retrieveFilterplace(); 
                            $data_place = array();
                            echo  "<select id='placeFilter' class='placeFilter' name='placeFilter'>";
                            while($row_place = mysql_fetch_assoc($something_place)){
                              array_push($data_place, $row_place);
                              echo  "<option value='".$row_place['placeNumber']."' >".$row_place['placeNumber']." </option>";
                              } 
                            echo "</select>";

                        if((isset($_POST['dateFilter'])) && (isset($_POST['placeFilter']))){
                            $result = mysql_query("SELECT nameNumber,time,user_id * FROM table_something WHERE '".$_POST['dateFilter']."' AND '".$_POST['placeFilter']."'   ");
                            echo "<option value=''>Select Your Album</option>";
                            while($row = mysql_fetch_array($result)){
                                echo "<option value='" . $row['nameNumber'] . "'>" . $row['time'] . "</option>";
                            }           

                        }

                          ?>
                          <div id="ajax_reply_div">
                         </div>
                         <input type="submit" value="Submit">
                        </div>
                </form>`

错误在此行中（$ row = mysql_fetch_array（$ result））{。过程是用户将从两个选择选项中进行选择，然后单击提交按钮，然后将从两个选择选项中输出与选择用户相关的行。任何帮助将非常感激。

Answer 1

您收到此错误是因为$result是布尔值。您可以阅读有关该功能如何工作的文档here on the PHP documentation site。

$result = mysql_query("SELECT nameNumber,time,user_id * FROM table_something WHERE '".$_POST['dateFilter']."' AND '".$_POST['placeFilter']."'   ");

您可能会收到此错误，因为您的SELECT查询形成错误，而是生成错误。

您不仅需要删除*，如下所示，而且您的WHERE子句可能无效，因为您实际上并未比较任何值。阅读有关WHERE子句here

的更多信息

$result = mysql_query("SELECT nameNumber,time,user_id FROM table_something WHERE '".$_POST['dateFilter']."' AND '".$_POST['placeFilter']."'   ");

您可能还想查看用于操作数据库的PDO或MySQLi扩展，因为mysql_*函数现已弃用并将被删除（请参阅php manual page?上的大红框）。另请参阅如何防止SQL Injection使代码更安全

无法输出与两个选择选项相关的数据库中的行

1 个答案: