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我sqli对致命错误的约束。
原标题:mysqli bind_param() fatal error
  • 时间:2011-07-05 12:18:12
  •  标签:
  • php
  • mysqli

我的《法典》中一位傲慢者能否帮助我?

<?php
  $db = new mysqli("localhost","root","","karmintalender");

  $owner_ID = 1;

  $sql = "SELECT name, kalender_ID FROM kalender WHERE ersteller_ID = ?";
  $stmt = $db->prepare($sql);
  $stmt->bind_param("i", $owner_ID);
  $stmt->execute();
  $stmt->bind_results($name, $kalender_ID);

  while ($stmt->fetch()) {
    echo $name . " " . $kalender_ID;
  }
?>

当我开场白时,这一错误似乎是“塔里差错:在G:xampphtdocs Karmintalender est.php第8行,要求成员履行对非目标的义务。

最佳回答

您在这方面的一个领域是 t的。

$sql = "SELECT name, kalender_ID FROM kalender WHERE ersteller_ID = ?";

此外,你还应检查价。

$db = new mysqli("localhost","root","","karmintalender");

 $owner_ID = 1;

 $sql = "SELECT name, kalender_ID FROM kalender WHERE ersteller_ID = ?";
 $stmt = $db->prepare($sql);
 if($stmt){
     $stmt->bind_param("i", $owner_ID);
     $stmt->execute();
     $stmt->bind_results($name, $kalender_ID);

     while ($stmt->fetch()) {
       echo $name . " " . $kalender_ID;
     }
 }
问题回答

它应为:stmt->bind_result($,kalender_ID);

下降





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