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Simple PHP Array Problem - IF EXIT

开发者 https://www.devze.com 2022-12-23 00:22 出处:网络
how can i convert this into an array? if someone se开发者_高级运维arches for \"lo\" he gets the text \"no query\", but how can i do this for more words? i tried it with array(\'1\',\'2\')..

how can i convert this into an array? if someone se开发者_高级运维arches for "lo" he gets the text "no query", but how can i do this for more words? i tried it with array('1','2')..

if ($query == 'lo')
{
  exit ('No Query.');
}

i want something like this

if ($query == 'lo', 'mip', 'get')
{
  exit ('No Query.');
}

so, if someone types mip he gets the message.. thank you!!


if ($query == 'lo' || $query == 'mip' || $query == 'get') {
    exit('No query');
}

Or if you might want to add many strings to check for, try using an array.

$bad_words = array('lo', 'mip', 'get', ... );
if (in_array($query, $bad_words)) {
    exit('No query');
}

Then adding new 'bad words' is as easy as adding them to the array.


Your array should look like this:

$array = array('lo', 'mip', 'get');

Now Search

if (in_array('mip', $array))
{
  echo 'We have a match';
}
else
{
  echo 'No match';
}
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