php - mysql_fetch_assoc "or die"语句已触发但仍在工作

Question

如果这是一个愚蠢的问题，请原谅我。我才开始编程一周...

我在第 87 行触发了获取错误“or die”语句(至少在我的编辑器中是 87，不确定它会如何显示在此处)。奇怪的是它仍然有效。接下来两行的变量正在正确设置和回显。任何关于为什么会发生这种情况的想法将不胜感激。谢谢。

<?php
include 'header.php';

//$results_per_page=10;
//$count_results=mysql_query("SELECT COUNT('zipcodes') FROM jobposts");
//$page_results=$count_results;
//echo $count_results;

//if ($count_results%$results_per_page!=0)
//{
//$page_results+=1;
//}


 /////// zip code search setup ///////////
 $homezip=22102;
 $radius=10;
 $id=1;

$query="select longitude,latitude from zipcodes where zipcode='$homezip'";
$run_query=mysql_query($query) or die ("run query error");
$findzip2=mysql_fetch_array($run_query) or die ("test");
$homelong=$findzip2['longitude'] or die ("array error");
$homelat=$findzip2['latitude'] or die ("array error");


function calcDist($lat_A, $long_A, $lat_B, $long_B) {

 $distance = sin(deg2rad($lat_A))
            * sin(deg2rad($lat_B))
            + cos(deg2rad($lat_A))
            * cos(deg2rad($lat_B))
            * cos(deg2rad($long_A - $long_B));

 $distance = (rad2deg(acos($distance))) * 69.09;

  return $distance;
  }
 //////// end zip code search setup ////////////////



 echo "<table border='1'>";

  if (isset($_POST['search']))
 {
  $keywords=$_POST['keywords'];
  $zipcode=$_POST['zipcode'];

  if ($zipcode=="")
  {
   ($zipcode_search="");
  }

   else
  {
   ($zipcode_search=" and zipcode='$zipcode'");
   }

   if (!(isset($_POST['hidden'])))
    {
    $search = "select * from jobposts where description LIKE '%$keywords%'            $zipcode_search";
    $run_search=mysql_query($search) or die ("cannot run search");


    //$zip_array=$main_array[''];
    //echo count($zip_array)."<br>";
    //echo $zip_array['zipcode'];


 ////store results in array
  ////pull zip codes out of array one at a time
  ////put the zip codes in the radius function - if statement
  ////if less than allowable radius - publish results

  $count=0;
    if (mysql_num_rows($run_search)>=1)
        {
        while ($query_rows=mysql_fetch_assoc($run_search))
            {
            $zip_array[$count]=($query_rows['zipcode']);

            $zip_to_test=$zip_array[$count];

            $testquery="select longitude,latitude from zipcodes where zipcode='$zip_to_test'";
            $testrun_query=mysql_query($testquery) or die ("run query error");
            $testfindzip2=mysql_fetch_assoc ($testrun_query) or die ("fetch error - testfindzip2");
            $testlong=$testfindzip2['longitude'] or die ("array error");
            $testlat=$testfindzip2['latitude'] or die ("array error");
            $count++;
            echo $testlong."<br>";
            if (calcDist($homelat,$homelong,$testlat,$testlong)<$radius)
            {
            echo "<tr><td>".$query_rows['company']."</td><td>".$query_rows['zipcode']."</td></tr>";
            }
            }
        }
        else
        {
        echo "No results found.";
        }
    }
    else
        {
        echo "Please submit search data before submitting.";
        }


    }
   else

   {
   echo "Failed to submit search request.";
  }



  ?>

Answer 1

mysql 函数仅返回 false，如果查询出现问题:语法错误、未连接、权限被拒绝等，这将触发您的 die()...

不返回结果的查询不是错误条件，并且不会触发或 die()。空结果集是完全有效的结果集。

还有，

$var = $othervar or die();

毫无意义，因为分配总是会成功(除非你耗尽内存或其他什么)。