Question

我正在处理我的网络项目，我正在尝试使用php和MySql数据库创建一个搜索栏。但每次我收到警告信息时，我已经尝试了Stack Overflow上的几乎所有解决方案。

但是任何一个都不会给我任何改进。在我的数据库中有一个名为'project_2013'的表，它有4列'ID，NAME，PROJECT，TITLE'。这是我用过的php代码。

    <?php
    if(isset($_POST['search'])){
        $searchTerm = $_POST['search'];
'%$searchTerm%'";
        $query = "SELECT * FORM project_2013 WHERE TITLE LIKE 
'%".mysqli_real_escape_string($con, $searchTerm)."%'";
        $result = mysqli_query($con,$query);
        $count = mysqli_num_rows($result);
        if ($count == 0){
            echo 'SORRY Nothing in our Server...';
        }
        else{
             while($row = mysqli_fetch_array($query)){
               $title = $row['TITLE'];
               $projectName = $row['PROJECT'];
               $name = $row['NAME'];
               echo '<br>';
               echo '<div class = "row">';//row start
               echo '<div id = "content" class = "col-lg-12 col-md-12 col-sm-12 col-xs-12">'; //div1
               echo '<b>Title:</b> ' .  $row['TITLE'] . '<br>';
               echo '<b>Research Paper:</b> ' . '<a href = "documents/2013_2014/' . $row['PROJECT'] .'" target = "_blank">'. $row['TITLE']  . '</a>' . '<br>';
              echo '<b>Conducted By:</b> ' . $row['NAME'] . '<br>';
              echo '</div>';//end of div1  
              echo '</div>';//row end
              echo '<hr>';
            }
        }
    }?>

这是表格标签部分。

<form action = "Home.php" method = "post">
    <div class="input-group">
    <input type="text" class="form-control" name = "search" placeholder="Search for Research...">
    <span class="input-group-btn">
    <button class="btn btn-default" type="submit" name="submit">
    <span class="glyphicon glyphicon-search"></span></button>
    </span>
    </div><!-- /input-group -->

Answer 1

使用：

"SELECT * FROM"

在SQL而不是：

"SELECT * FORM

注意：

在将SQL添加到代码中之前，您应该验证它。

不应在此处发布带语法问题的代码。

Answer 2

首先，您应该更改该查询部分中的＆＃34; SELECT * FROM＆＃34; 。

部分还有一个人认为你应该改变    mysqli_fetch_array（$ query）而不是 mysqli_fetch_array（$ result）。

检查下面正确的代码

<?php
if(isset($_POST['search'])){
    $searchTerm = $_POST['search'];
    //$searchTerm = preg_replace("#[^0-9 a-z]#i","",$searchTerm);
    //$query = "SELECT * FROM project_2013 WHERE TITLE LIKE '%$searchTerm%'";
    $query = "SELECT * FORM project_2013 WHERE TITLE LIKE '%".mysqli_real_escape_string($con, $searchTerm)."%'";
    $result = mysqli_query($con,$query);
    $count = mysqli_num_rows($result);
    if ($count == 0){
        echo 'SORRY Nothing in our Server...';
    }
    else{
            while($row = mysqli_fetch_array($result)){
                $title = $row['TITLE'];
                $projectName = $row['PROJECT'];
                $name = $row['NAME'];

                echo '<br>';
                echo '<div class = "row">';//row start
                    echo '<div id = "content" class = "col-lg-12 col-md-12 col-sm-12 col-xs-12">'; //div1
                        echo '<b>Title:</b> ' .  $row['TITLE'] . '<br>';
                        echo '<b>Research Paper:</b> ' . '<a href = "documents/2013_2014/' . $row['PROJECT'] .'" target = "_blank">'. $row['TITLE']  . '</a>' . '<br>';
                        echo '<b>Conducted By:</b> ' . $row['NAME'] . '<br>';
                    echo '</div>';//end of div1  
                echo '</div>';//row end
                echo '<hr>';
        }
    }
}?>

警告：mysqli_num_rows（）期望参数1为mysqli_result，

2 个答案: