簡體   English   中英

mysqli_fetch_assoc無法正常工作

[英]mysqli_fetch_assoc not work

為什么第二次嘗試時第一行沒有顯示? 在屏幕截圖中,第一次回顯時突出顯示第一行。

輸出

[![enter image description here][1]][1]/*=================================================================================================================
        mysqli_fetch_assoc - returns data in an associative array
        mysqli_fetch_row - returns data in a numeric array
        mysqli_fetch_array - returns data in both an associative and a numeric array
    */
    $query2 = "SELECT * FROM User";
    //$result3 = mysqli_query ( $con, $query2) or die ("Couldn't execute SELECT query");
    $result3 = mysqli_query ( $con, $query2) or die (mysqli_error($con));

    $row = mysqli_fetch_assoc($result3); //returns an array called $row with column names as keys. Get one row of data
                                         //If you need more than one row of data, use the function in a loop

    echo $row['Title'] ." - ". $row['FName'] ." - ". $row['LName']; //one row of data. It is fine when checking a password

    //extract function - splits the array into variables that have same name as the key
    extract($row);
    echo "<hr>".$FName; //variable having same name as the key, also identical to column name
    echo "<br>Number of rows in the User table: ".mysqli_num_rows($result3);

    echo "<table>";
    echo "<th>User Id</th> <th>Title</th> <th>First Name</th> <th>Last Name</th> <th>Email</th>";
    while( $row = mysqli_fetch_assoc($result3) ){
        extract($row);
        echo "<tr>";
        echo "<td>".$UserId."</td>" . "<td>".$Title."</td>" . "<td>".$FName."</td>" . "<td>".$LName."</td>" . "<td>".$Email."</td>" ;
        echo "</tr>";
    }
    echo "</table>";

    echo "<hr>My connection to the MySQL database";
$row = mysqli_fetch_assoc($result3); 

因為當您在循環之前執行此操作時,會將指針移到下一行,因此循環從第二行開始。

在循環之前,您可以將指針重置為第一行

mysqli_data_seek($result3,0);

因為,您兩次執行mysqli_fetch_assoc,而沒有重新設置指針。 Hanky Panky說的沒錯。

暫無
暫無

聲明:本站的技術帖子網頁,遵循CC BY-SA 4.0協議,如果您需要轉載,請注明本站網址或者原文地址。任何問題請咨詢:yoyou2525@163.com.

 
粵ICP備18138465號  © 2020-2024 STACKOOM.COM