Question

我遇到的问题是当我回显或打印以下变量时，我收到的数据只是我表格中列出的最后一项业务的数据。

目前无论我在哪个列表中点击，我都会获得返回上一次业务的相同数据集。

正如您在下面的代码中所看到的，我将从点击的列表中传递business_name以在我的查询中使用，以查找相关的业务配置文件信息。

$business_name = mysql_real_escape_string($_GET['business_name']);

$query = "SELECT
business_id,
category,
years_recommended,
profile_size,
business_name,
established,
employees,
service,
strengths,
ideal_for,
reassurance

FROM
business_data

WHERE
business_name = '$business_name'

AND
profile_size = 'A'

OR
profile_size = 'B'

OR
profile_size = 'C'

OR
profile_size = 'D'

OR
profile_size = 'E'";

$result = mysql_query($query, $dbc)
or die (mysql_error($dbc));

while($row = mysql_fetch_array($result)) {

$business_id = $row["business_id"];
$profile_size = $row["profile_size"];
$category = $row["category"];
$years = $row["years_recommended"];
$established = $row["established"];
$employees = $row["employees"];
$service = $row["service"];
$strengths = $row["strengths"];
$ideal_for = $row["ideal_for"];
$reassurance = $row["reassurance"];
}

echo...

如果您需要更多信息，请告诉我们。

我的代码有什么问题吗？

非常感谢提前。

Answer 1

您的echo调用位于获取循环之外，因此即使返回其他结果，您也只会看到最后的结果。

while($row = mysql_fetch_array($result)) {

  $business_id = $row["business_id"];
  $profile_size = $row["profile_size"];
  $category = $row["category"];
  $years = $row["years_recommended"];
  $established = $row["established"];
  $employees = $row["employees"];
  $service = $row["service"];
  $strengths = $row["strengths"];
  $ideal_for = $row["ideal_for"];
  $reassurance = $row["reassurance"];

  // Echo **inside** the loop
  echo...
}

如果您愿意，可以将所有结果存储在一个大型数组中，然后可以根据需要多次在脚本中的任何位置使用：

// Array for all results
$results = array();
while($row = mysql_fetch_array($result)) {
  // Append each row fetched onto the big array
  $results[] = $row;
}
// Now use it as needed:
foreach ($results as $r) {
  echo $r['profile_size'];
  print_r($r);
}

Answer 2

你的回音应该在循环中

为什么查询只返回一组数据？

2 个答案: