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      1. 帶有 mysqli 的 Bind_param 非對象錯誤

        Bind_param Non-Object Error w/ mysqli(帶有 mysqli 的 Bind_param 非對象錯誤)
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                1. 本文介紹了帶有 mysqli 的 Bind_param 非對象錯誤的處理方法,對大家解決問題具有一定的參考價值,需要的朋友們下面隨著小編來一起學習吧!

                  問題描述

                  嘗試為將跟蹤每日視圖的表插入初始行時,出現錯誤:

                  When attempting to insert the initial row for a table that will track daily views, I am getting the error:

                  致命錯誤:在 157 行/.../functions.php 中的非對象上調用成員函數 bind_param()強>

                  Fatal error: Call to a member function bind_param() on a non-object in /.../functions.php on line 157

                  該行是以下組的最后一行:

                  That line is the last of the following group:

                  if($stats_found) {
                   $sqlquery = "UPDATE vid_stats SET views = ? WHERE title = ? AND format = ? AND date = ? AND results = ?";
                   $views++;
                  } else {
                   $sqlquery = "INSERT INTO vid_stats (views, title, format, results) values (?, ?, ?, ? )";
                   $views = 1;
                  }
                  
                  $stmt = $mysqli->prepare($sqlquery);
                  /* bind parameters for markers */
                  $stmt->bind_param("dsss", $views, $title, $format, "success");
                  

                  關于問題的任何提示?

                  以防萬一這是周圍代碼的問題,這里是完整的功能:

                  Just in case it's an issue with the surrounding code, here is the complete function:

                  function updateViewCount($title, $format, $results) {
                   //update view count
                   global $mysqli;
                   $views = 0;
                   if ($stmt = $mysqli->prepare("SELECT views FROM vid_stats WHERE title = ? AND format = ? AND date = ?")) {
                  
                    /* bind parameters for markers */
                    $stmt->bind_param("ssd", $title, $format, date("Y-m-d"));
                  
                    /* execute query */
                    $stmt->execute();
                  
                    /* bind result variables */
                    $stmt->bind_result($views);
                  
                    /* fetch value */
                    if ($stmt->fetch()) {
                     $stats_found = true;
                    } else { $stats_found = false; }
                  
                    /* close statement */
                    $stmt->close();
                  
                    if($stats_found) {
                     $sqlquery = "UPDATE vid_stats SET views = ? WHERE title = ? AND format = ? AND date = ? AND results = ?";
                     $views++;
                    } else {
                     $sqlquery = "INSERT INTO vid_stats (views, title, format, results) values (?, ?, ?, ? )";
                     $views = 1;
                    }
                  
                    $stmt = $mysqli->prepare($sqlquery);
                    /* bind parameters for markers */
                    echo $sqlquery."<br>".$views."<br>".$title."<br>".$format;
                    $stmt->bind_param("dsss", $views, $title, $format, "success");
                  
                    /* execute query */
                    $stmt->execute();
                  
                    /* close statement */
                    $stmt->close();
                   }
                  }
                  

                  推薦答案

                  問題是用戶錯誤:我把 result 列的名稱弄錯了.

                  The problem was user error: I had the name of the result column wrong.

                  當我在行 $stmt = $mysqli->prepare($sqlquery); 之后添加 echo $mysqli->error; 時發現了這一點,它顯示列名錯誤.

                  This was uncovered when I added echo $mysqli->error; after the line $stmt = $mysqli->prepare($sqlquery); which revealed the column-name error.

                  這篇關于帶有 mysqli 的 Bind_param 非對象錯誤的文章就介紹到這了,希望我們推薦的答案對大家有所幫助,也希望大家多多支持html5模板網!

                  【網站聲明】本站部分內容來源于互聯網,旨在幫助大家更快的解決問題,如果有圖片或者內容侵犯了您的權益,請聯系我們刪除處理,感謝您的支持!

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