将值从 HTML 表单传输到 PHP 文件时，变量的值未正确存储

Question

谢谢 Ben，但现在我面临的另一个问题是，当我将分页添加到代码中时，第一次没有显示结果。请参阅下面代码的后半部分。请帮忙

if(isset($_GET['k'])){ $k1 = $_GET['k']; } else { $k1 = ''; }
echo $k1;
    $term = explode(" ", $k1);

$query = "SELECT * FROM database ";

foreach ($term as $each) 
{
    echo $each;
$i++;
    if($i==1)
    {
        $query .= "WHERE keywords LIKE '%$each%' ";
    }

else {
    $query .= "OR WHERE keywords LIKE '%$each%' ";
}

}

$per_pages=3;
$page_query = mysql_query("SELECT COUNT('title') FROM kcpdatabase");
    $pages = ceil(mysql_result($page_query, 0)/$per_pages)  or die

       ($page_query."<br/><br/>".mysql_error());    
       $page = (isset($_GET['page'])) ? (int) 
       ($_GET['page']) : 1;                             
$start = ($page - 1) * $per_pages;  
$query .= "LIMIT $start, $per_pages";


$ourquery1 = mysql_query ($query);
if(!$ourquery1)
echo "No query found";
$row1 = mysql_num_rows ($ourquery1);

if($pages >= 1 && $page <= $pages){ 

    for($x = 1; $x <= $pages; $x++)
    {

        echo '<a href="?page='.$x.'">'.$x.'</a> ';
    }

if ($row1 > 0)
{

    while($result = mysql_fetch_assoc($ourquery1))
    {
           echo "<tr>";
        echo "<td>";
        $title = $result['title'];
        $link = $result['link'];
        $region = $result['region'];
        $sector = $result['sector'];
        $theme = $result['theme'];      
        echo "<td> <a href=$link><h3>$title<h3></a>";
        echo "<h4>Sector: $sector <br>Theme: $theme <br> Region: $region  
                          </td>  </tr>";
    }
}   

}
echo "</tbody>";

下面是我尝试搜索在文本框中输入的单词/短语的部分代码。当我使用“$k1 = isset($_GET['k']);”从 php 文件中的表单中捕获值时存储在变量“$each”中的值是“1”，而不是用户输入的单词或短语。这会弄乱正在执行搜索功能的查询。请帮我定位错误。

请注意，“k”是文本框的名称，定义在下面的表单代码中。

    <form name="keywordquery" method="get" action="page2.php">
<fieldset class="fieldsetclass"><legend class="legendclass">Search by Keywords</legend>
      <div id="searchbox">
<input type="text" name="k" value="<?php if(isset($_GET['k'])){echo htmlentities($_GET
       ['k']);} ?>" style="border: 1px, thin; width:92%; "/> 
<input type="image" style="margin-bottom: 0; margin-top: 2px;" src="search.png"
      value="submit" />
</div>
</fieldset>
    </form>
    </div>
<table cellpadding="0" cellspacing="0" border="1">

    <tbody>    
<?php

$connection = mysql_connect('', '', '');
if(!$connection)
echo "No database connected";
$dbase = mysql_select_db("", $connection);
if(!$dbase)
echo "No datatable connected";



$k1 = isset($_GET['k']);
echo $k1;
$term = explode(" ", $k1);

$query = "SELECT * FROM datatable ";

foreach ($term as $each) 
{
    echo $each;
$i++;
    if($i==1)
    {
        $query .= "WHERE keywords LIKE '%$each%' ";
    }

else {
    $query .= "OR WHERE keywords LIKE '%$each%' ";
}

}

Answer 1

$k1 = isset($_GET['k']);

将 $k1 设置为 1，因为它正在检查是否已设置 - 在本例中已设置，isset() 返回 true 或 1。

你想要的是：

if(isset($_GET['k'])){ $k1 = $_GET['k']; } else { $k1 = ''; }

或类似的。

Answer 2

那是因为 isset() 是一个布尔函数，请查看以下链接http://php.net/manual/en/function.isset.php

就这样做

if(isset($_GET['k']))
    $k1 = $_GET['k'];