Question

我试图用我所有的产品回应产品清单，但我在mysql中有3个产品。但是由于某些原因，当我回复我的列表时，它只显示了一个产品，第一个列表。虽然我想看到所有的产品（3）。

如果这可能与我正在使用相关：服务器版本：5.5.36 - MySQL 和[PHP：5.4.27]

    // This block grabs the whole list for viewing
include "../sitescripts/connect_to_mysql.php";
    $product_list = "";
    $sql = mysql_query("SELECT * FROM products ORDER BY date_added DESC");
    $productCount = mysql_num_rows($sql); // count the output amount
    if ($productCount > 0) {
        while($row = mysql_fetch_array($sql)){ 
                 $id = $row["id"];
                 $product_name = $row["product_name"];
                 $price = $row["price"];
                 $date_added = strftime("%b %d, %Y", strtotime($row["date_added"]));
                 $product_list = "Product ID: $id - <strong>$product_name</strong> - $$price - <em>Added $date_added</em> &nbsp; &nbsp; &nbsp; <a href='inventory_edit.php?pid=$id'>edit</a> &bull; <a href='inventory_list.php?deleteid=$id'>delete</a><br />";
        }
    } else {
        $product_list = "You have no products listed in your store yet";
    }

Answer 1

您的商品列表仅显示您行中的最后一项。

$product_list = "Product ID: $id - <strong>$product_name</strong> - $$price - <em>Added $date_added</em> &nbsp; &nbsp; &nbsp; <a href='inventory_edit.php?pid=$id'>edit</a> &bull; <a href='inventory_list.php?deleteid=$id'>delete</a><br />";

正在覆盖每次迭代的值，我认为你打算这样做：

$product_list .= "Product ID: $id - <strong>$product_name</strong> - $$price - <em>Added $date_added</em> &nbsp; &nbsp; &nbsp; <a href='inventory_edit.php?pid=$id'>edit</a> &bull; <a href='inventory_list.php?deleteid=$id'>delete</a><br />";

用于在每次迭代时附加字符串值。请注意=符号前的句号。

Answer 2

注意：

您应该回复{strong} IF的$product_list变量 INSIDE ELSE 条件。 NOT AFTER 条件。仅拥有最后一行的原因是因为最后一行行将覆盖循环的前一行。
我还建议使用 MySQLi 而不是弃用 MySQL 。
在使用变量之前，至少使用 mysqli_real_escape_string 函数来阻止SQL injections。

如果您将代码转换为 MySQLi ，它将如下所示：

<强> connect_to_mysql.php ：

<?php

/* ESTABLISH CONNECTION */

$con=mysqli_connect("YourHost","YourUsername","YourPassword","YourDatabase"); /* REPLACE NECESSARY DATA */

if(mysqli_connect_errno()){

echo "Error".mysqli_connect_error();
}

?>

您的抓取/主文件：

    include "../sitescripts/connect_to_mysql.php"; /* INCLUDE CONNECTION */
    $product_list = "";
    $sql = mysqli_query($con,"SELECT * FROM products ORDER BY date_added DESC");
    $productCount = mysqli_num_rows($sql); // count the output amount
    if ($productCount > 0) {
        while($row = mysqli_fetch_array($sql)){ 

                 /* USED mysqli_real_escape_string FUNCTION TO PREVENT SQL INJECTION */

                 $id = mysqli_real_escape_string($con,$row["id"]);
                 $product_name = mysqli_real_escape_string($con,$row["product_name"]);
                 $price = mysqli_real_escape_string($con,$row["price"]);
                 $date_added = strftime("%b %d, %Y", strtotime($row["date_added"]));
                 $product_list = "Product ID: ".$id." - <strong>".$product_name."</strong> - $".$price." - <em>Added ".$date_added."</em> &nbsp; &nbsp; &nbsp; <a href='inventory_edit.php?pid=$id'>edit</a> &bull; <a href='inventory_list.php?deleteid=$id'>delete</a><br />";

                 echo $product_list; /* PRINT $product_list VARIABLE */

        } /* END OF WHILE LOOP */

    } /* END OF $productCount VARIABLE IS MORE THAN 0 */

    else {
        $product_list = "You have no products listed in your store yet";
        echo $product_list; /* PRINT $product_list VARIABLE */
    } /* END OF ELSE */

    /* I ASSUME YOU TRY TO PRINT THE $product_list VARIABLE IN HERE */

PHP为什么我的while循环回声只有一个项目而我在mysql中有3个项目

2 个答案: