Php图像显示错误

时间:2014-08-22 04:55:19

标签: php

当我运行此查询时,图像正在从PHP正确显示。

<?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']." />";
}
?>

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();
}
?>

当我在上面运行相同的代码时,只需对搜索脚本进行一些更改:

<?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']." />";
}
?>

它显示错误,未定义索引:pagecontent,也没有显示图像。请提前帮助和谢谢。

2 个答案:

答案 0 :(得分:3)

在您的查询中

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

您没有获取列pagecontent并尝试在

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

将您的查询更改为

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

答案 1 :(得分:1)

尝试并从$ sql获取打印值..您将了解如何返回它..

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