Question

问题

编译完这段代码后，我在显示页面时出现问题，但我看不出它有什么问题，因为它没有出现在网上而无法调试。

PHP和PostgreSQL代码：

<?php       
    // Connecting, selecting database
    $dbconn = pg_connect("host=*****  port=*****
    dbname=***** user=***** password=*****")
    or die('Could not connect: ' . pg_last_error());

    //collect
    if(isset($_POST['search'])) {
        $searchq = $_POST['search'];
        $searchq = preg_replace("#[^0-9a-z]#i"."".$searchq);

    // Performing SQL query
    $query = "SELECT * FROM enumbers WHERE code LIKE '%$searchq%'") or die ("could not search!");
       $result = query($query);
    if($result = 0){
        $output = 'There is no such E-Number!'
    }else{
        while($row = mysql_fetch_array($query)) {
            $code = $row['code'];
            $name = $row['name'];
            $type = $row['type'];
            $vegan = $row['vegan'];

            $output .= '<div> '.vegan.' ';
            }
        }
    }
?>

表格和印刷代码：

 <div id="tablebox">
    <!-- Search bar -->
    <p>Is It Vegan?</p>
    <form name="form1" method="post" action="searchEnumbers.php">
        <input name="search" type="text" size="30" maxlength="5" />
        <input name="submit" type="submit" value="Search" />
    </form>

    <?php
        print("$output");
    ?>
</div>

更新

我认为问题可能是我使用了一些MySQL代码，但我不知道我是不是。

Answer 1

您正在使用mysql_fetch_array而不是postgres方法。请参阅样本

    <?php 
    // Connecting, selecting database
    $dbconn = pg_connect("host=****  port=****
    dbname=**** user=**** password=****")
    or die('Could not connect: ' . pg_last_error());
     $output = '';

    //collect
    if(isset($_POST['search'])) {
        $searchq = $_POST['search'];
        // $searchq = preg_replace("#[^0-9a-z]#i"."".$searchq);

    // Performing SQL query
    $query = "SELECT * FROM enumbers WHERE code LIKE '%$searchq%'";
    $ret = pg_query($dbconn, $query);
   if(!$ret){
      echo pg_last_error($dbconn);
      exit;
   } 
   $output = '';
   while($row = pg_fetch_assoc($ret)){
            $code = $row['code'];
            print_r($row);
            $name = $row['name'];
            $type = $row['type'];
            $vegan = $row['vegan'];

            $output .= '<div> '.vegan.' ';
   }
}
   echo "Operation done successfully\n";
   pg_close($dbconn);

?>

编译PHP和PostgreSQL后，页面空白

1 个答案: