我使用简单的html dom将网站中的数据抓取到我的数据库中并显示在我的网页上。但每次运行该文件时,重复数据也会插入到数据库中。如何检查数据是否已存在于数据库中?这是我的抓取文件:
<?php
$con=mysqli_connect("localhost","root","","crawling");\
mysql_connect("localhost", "root", "")or die("cannot connect");
mysql_select_db("crawling")or die("cannot select DB");
include "domcrawl.php";
$url="http://www.bgr.in/category/reviews/";
$html=file_get_html($url);
//$arr=$html->find('table[class=findList] tbody tr td[class=result_text]');
$m=$html->find('img');
$b=$html->find('a');
$c=$html->find('p');
$imghead = $b[21]->innertext;
$img = $m[3];
$imgtext = $c[0];
$sql = sprintf("INSERT INTO image1
( head, image, text, name)
VALUES
( '%s', '%s', '%s', '%s')",
mysql_real_escape_string($imghead),
mysql_real_escape_string($img),
mysql_real_escape_string($imgtext),
mysql_real_escape_string("gm")
);
mysql_query($sql);
$sql = "SELECT head FROM image1 WHERE name='gm'";
$sql1 = "SELECT image FROM image1 WHERE name='gm'";
$sql2 = "SELECT text FROM image1 WHERE name='gm'";
$result = mysql_query("$sql");
$result1 = mysql_query("$sql1");
$result2 = mysql_query("$sql2");
$head_get= mysql_result($result, 0);
$img_get= mysql_result($result1, 0);
$text_get= mysql_result($result2, 0);
echo "<br><br>";
echo $head_get;
echo "<br><br>";
echo $img_get;
echo $text_get;
?>
答案 0 :(得分:1)
在获取对象属性之前必须检查,在您的情况下,它正在查找空对象
$link = $node->getElementsByTagName('link')->item(0);
if(!empty($link)){
$nodeValue = $link->nodeValue,
}
'link' => $nodeValue;
同样适用于所有
答案 1 :(得分:0)
假设'date' => $node->getElementsByTagName('pubDate')->item(0)->nodeValue
是第11行,似乎没有标记为pubDate
的元素,这就是$node->getElementsByTagName('pubDate')->item(0)
返回null
或false
的原因。< / p>