Question

我有下面的代码并且它应该正常工作但是不是将结果回显到屏幕，我需要将结果存储在名为$ var的php变量中。我该怎么做呢？

<?php
$sql = "SELECT id_member FROM smf_members WHERE FIND_IN_SET(24,additional_groups)";

$con = mysql_connect('localhost', 'sqluser', 'sqlpass');
if (!$con) {
   die('Could not connect: ' . mysql_error());
}

mysql_select_db("test_db");
$result = mysql_query($sql, $con);

while ($row = mysql_fetch_array($result)) {
   echo $row['id_member'];
}

mysql_close($con);
?>

Answer 1

根据您想要实现的目标，有几种可能的方法

$var = "";

while ($row = mysql_fetch_array($result)) {
  $var .= $row['id_member'] . "\n";
}

$var = array();

while ($row = mysql_fetch_array($result)) {
  $var[] = $row['id_member'];
}

Answer 2

将echo替换为$var[]。

这会将每个结果推送到数组的末尾。首先定义变量可能会很好。

$var = array();
while ($row = mysql_fetch_array($result)) {

  $var[] = $row['id_member'];
}

Answer 3

<?php
$sql = "SELECT id_member FROM smf_members WHERE FIND_IN_SET(24,additional_groups)";

$con = mysql_connect('localhost', 'sqluser', 'sqlpass');
if (!$con) {
    die('Could not connect: ' . mysql_error());
}

mysql_select_db("test_db");

$result = mysql_query($sql, $con);

$v = array(); // $v instead of $var, since var is a keyword and may cause troubles

while ($row = mysql_fetch_array($result)) {
array_push($v, $row['id_member']); 
// or
//$v[] = $row['id_member'];
}

mysql_close($con);

＆GT;

Answer 4

如果select语句将返回多个结果，则需要将值存储在数组中：

$member_ids = array();
while ($row = mysql_fetch_array($result)) {
  $member_ids[] = $row['id_member'];
}

如果select语句将返回单个结果（您可以通过将LIMIT 1附加到$sql变量的值来确保。

$row = mysql_fetch_array($result);
$member_id = $row['id_member'];

在php变量中存储SQL调用的结果

4 个答案: