插入语句在foreach语句中不起作用

Question

I have a crawler which crawls the sites that begin with www.bbc.co.uk/news. It it grabs all the links that start with http://www.bbc.co.uk/news and finds their description, link and title and inserts them into a database.

For some reason, it doesn't seem to be inserting.

Any ideaS?

PS There is completely not output, completely blank web page

   foreach ($links as $link) {
    $output = array(
"title"       => Titles($link), //dont know what Titles is, variable or string?
"description" => getMetas($link),
"keywords" => getKeywords($link), 
"link"        => $link                 
 );
if (empty($output["description"])) {
$output["description"] = getWord($link);
 }

 if (substr($ouput, 0, 26) == "http://www.bbc.co.uk/news/") {

 $data = '"' . implode('" , "', $output) . '"';
 $success = mysql_query( "INSERT INTO news_story (`title`, `description` , `keywords`, `link`)
 VALUES (" . $data . ")") or zerror_reporting();
 if ($sucess) {
echo "YEAH!";
   }

   if (!$sucess) {
echo "NO!!";
    }
    print_r($data);
     }}

Answer 1

Problem is Here:

 if (substr($ouput, 0, 26) == "http://www.bbc.co.uk/news/") {

   $data = '"' . implode('" , "', $output) . '"';
  $success = mysql_query( "INSERT INTO news_story (`title`, `description` , `keywords`, `link`)
  VALUES (" . $data . ")") or zerror_reporting();
 if ($sucess) {
echo "YEAH!";
  }

Where is your $ouput variable...I think you wanted to write $output ..but it also didnot execute because $output variable is a array not a string

Answer 2

The blank white page is a PHP Fatal error that produces a 500 Internal Server Error response. That is caused by this undefined function zerror_reporting() :

mysql_query(...) or zerror_reporting();

Change that to something like

mysql_query(...) or trigger_error(mysql_error());

The trigger_error() call will add the mysql error to your error log.

The second problem is you're trying to substr() on an array, you should be doing that on the link element:

 if (substr($output['link'], 0, 26) == "http://www.bbc.co.uk/news/") {

Answer 3

在插入数据库之前对您的值进行消毒

Answer 4

Solution by @Mrinmoy is correct but there seems more issues in the code since your code never touched this further.

First set display errors:

ini_set('error_reporting',E_ALL);
ini_set('display_errors','on');
foreach ($links as $link) {

PHP talks a lot, if you can listen. I personally use E_ALL|E_STRICT but thats little too much for today. :) Then sanitize your data or you would rarely succeed in inserting records. Your data will have lot of sentences:

 $output = array(
"title"       => mysql_real_escape_string(Titles($link)), //dont know what Titles is, variable or string?
"description" => mysql_real_escape_string(getMetas($link)),
"keywords" => mysql_real_escape_string(getKeywords($link)), 
"link"        => mysql_real_escape_string($link)                 
 );
if (empty($output["description"])) {
$output["description"] = mysql_real_escape_string(getWord($link));
 }

Then correct the variable typo and use the link index of the output array:

if (substr($output['link'], 0, 26) == "http://www.bbc.co.uk/news/") {

At the end, if you still don't get data, you definitely will know much more to fix it yourself. And use print_r($output); echo $data; print_r($output); echo $data; before calling mysql_query. Another way to track progress by filling code with echo __LINE__ . "\\n"; echo __LINE__ . "\\n"; to see where it died. Verify there is a method in your code by the name zerror_reporting or replace with die(mysql_error());