显示Php图像错误

Question

当我运行这个查询时，图像将正确地显示在php中。

<?php
mysql_connect("localhost", "root", "");
mysql_select_db("dawat");
$submit=$_GET['str'] ;
$sql = mysql_query("SELECT * FROM searchengine WHERE pagecontent LIKE '%$_GET%' ");    
while($row=mysql_fetch_array($sql)) {
echo "<img src=image.php?pagecontent=".$row['pagecontent']." />";
}
?>

以及image.php的代码：

<?php
ini_set('display_errors',1);
error_reporting(E_ALL);
$conn = mysql_connect("localhost","root","");
if(!$conn)
{
echo mysql_error();
}
$db = mysql_select_db("dawat",$conn);
if(!$db)
{
echo mysql_error();
}
$pagecontent = $_GET['pagecontent'];
$q = "SELECT pageurl FROM searchengine where pagecontent='$pagecontent'";
$sql = mysql_query("$q",$conn);
if($sql)
{
$row = mysql_fetch_array($sql);
echo $row['pageurl'];
}
else
{
echo mysql_error();
}
?>

当我运行上述相同的代码时，搜索脚本中的几处更改：

<?php
mysql_connect("localhost", "root", "");
mysql_select_db("dawat");
$submit=$_GET['str'] ;
$sql = mysql_query("SELECT pageurl, BIT_COUNT(pagecontent^'$submit') 
FROM searchengine WHERE pagecontent < 20 ORDER BY pagecontent ASC;
") or die(mysql_error());
while($row=mysql_fetch_array($sql)) {
echo "<img src=image.php?pagecontent=".$row['pagecontent']." />";
}
?>

它显示了未定义索引的错误:分页，也没有预先显示images.Please帮助和感谢。

Answer 1

在查询中

$sql = mysql_query("SELECT pageurl, BIT_COUNT(pagecontent^'$submit') 
FROM searchengine WHERE pagecontent < 20 ORDER BY pagecontent ASC;
") or die(mysql_error());

您没有获取列pagecontent，而是尝试在

echo "<img src=image.php?pagecontent=".$row['pagecontent']." />";
                                                   ^

所以把你的查询改为，

$sql = mysql_query("SELECT pageurl,pagecontent, BIT_COUNT(pagecontent^'$submit') 
FROM searchengine WHERE pagecontent < 20 ORDER BY pagecontent ASC;
") or die(mysql_error());

Answer 2

试一试，从$sql获取打印值。你会从如何归还它中得到一些想法。

$sql = mysql_query("SELECT pageurl,pagecontent, BIT_COUNT(pagecontent^'$submit') 
FROM searchengine WHERE pagecontent < 20 ORDER BY pagecontent ASC;
") or die(mysql_error());