如何在js中搜索Id

Question

我正在尝试从dataTable中选择行，但是在PHP下的js代码中出现了问题。下面给出了。下面我首先从数据库中获取数据，然后将它们添加到dataTable中。代码完美地工作到此处..但是当我试图从表格中选择任何一行时，它会显示错误说明

解析错误：语法错误，意外＆＃39; getval＆＃39; （T_STRING）在第43行的C：\ xampp \ htdocs \ Trying \ fetch.php

，var table = document.getElementById('getval');几乎就在代码的末尾。

`

    <?php
            $connect = mysqli_connect("localhost","root","","ajmal");
            $output = '';

            $sql = "SELECT                                              
            medicinName,pricerPerSheet,dealerID,availAbleAt,district,place 
            FROM medicinalinfo WHERE medicinName LIKE 
            '%".$_POST["search"]."%'";
            $result = mysqli_query($connect,$sql);


            if(mysqli_num_rows($result) > 0)
            {
                $output .= '<h4 align="center" class="h4_search">Search 
                Result</h4>';
                $output .= '<div class="row">';
                $output .= '<div class="col-md-8 col-md-offset-1 well">';
                $output .= '<div class="table-responsive">
                    <table id="tbl" class="table table-bordered table- 
                    striped table-hover">
                        <tr>
                            <th>Medicin Name</th>
                            <th>Price Per Sheet</th>
                            <th>Availble At</th>
                            <th>District</th>
                            <th>Area</th>
                        </tr>';
               $output .= '</div>';
               $output .= '</div>';
               $output .= '</div>';
               while ($row = mysqli_fetch_array($result)) {
               $output .= '
               <tbody id="getval">
                    <tr>
                        <td>'.$row['medicinName'].'</td>
                        <td>'.$row['pricerPerSheet'].'</td>
                        <td>'.$row['availAbleAt'].'</td>
                        <td>'.$row['district'].'</td>
                        <td>'.$row['place'].'</td>
                    </tr>
               </tbody>
             ';
             }
             $output.='</table>';

             $output.='<script>

             var table = document.getElementById('getval');

             for(var i=0; i<table.rows.length; i++){
                table.row[i].onclick = function(){
                    alert(this.cells[0].innerHTML);
                };
             }

             </script>
          ';
          echo $output;
          }
        else
        {
           echo '<h4 align="center" class="h4_search">Data Not Found</h4>';
        }


     ?>

`

Answer 1

您正在为该字符串使用单引号，因此您必须在该JS位中转义它们，或更改为双引号。

因此，该行应为：

var table = document.getElementById(\'getval\');

或

var table = document.getElementById("getval");

修改 好的，这是整个代码，更正了，你有几个错误：

<?php

$connect = mysqli_connect("localhost", "root", "", "ajmal");
$output = '';

$sql = "SELECT                                              
            medicinName,pricerPerSheet,dealerID,availAbleAt,district,place 
            FROM medicinalinfo WHERE medicinName LIKE 
            '%" . $_POST["search"] . "%'";
$result = mysqli_query($connect, $sql);


if (mysqli_num_rows($result) > 0) {
    $output .= '<h4 align="center" class="h4_search">Search
            Result</h4>';
    $output .= '<div class="row">';
    $output .= '<div class="col-md-8 col-md-offset-1 well">';
    $output .= '<div class="table-responsive">
                <table id="tbl" class="table table-bordered table-
                striped table-hover">
                    <tr>
                        <th>Medicin Name</th>
                        <th>Price Per Sheet</th>
                        <th>Availble At</th>
                        <th>District</th>
                        <th>Area</th>
                    </tr>
                    <tbody id="getval">';
    while ($row = mysqli_fetch_array($result)) {
        $output .= '
                    <tr>
                        <td>' . $row['medicinName'] . '</td>
                        <td>' . $row['pricerPerSheet'] . '</td>
                        <td>' . $row['availAbleAt'] . '</td>
                        <td>' . $row['district'] . '</td>
                        <td>' . $row['place'] . '</td>
                    </tr>';
    }
    $output .= '</tbody></table>';
    $output .= '</div>';
    $output .= '</div>';
    $output .= '</div>';
    $output .= '<script>
             var table = document.getElementById("getval");
             for(var i=0; i<table.rows.length; i++){
                table.rows[i].onclick = function(){
                    alert(this.cells[0].innerHTML);
                };
             }

             </script>
          ';
    echo $output;
} else {
    echo '<h4 align="center" class="h4_search">Data Not Found</h4>';
}
?>

首先，你在表格中有关闭div，这是错误的，我在你的代码中纠正了这一点。然后，你有多个具有相同id的tbody元素，这就是造成混乱的原因。在mysqli_fetch_array循环之外移动了tbody段，因此我们只有一个tbody和多个行。此外，在javascript部分，它应该是

table.rows[i].onclick

而不是

table.row[i].onclick