而不是一次在mysql DB中搜索一个变量，而是我想搜索多个变量，并将其放入初始查询字符串中
所以而不是
$query="SELECT * FROM table1 WHERE state = 'CA' OR state = 'CO' OR state = 'TX'";
我要
$states = ("CA,CO,TX");
$query="SELECT * FROM table1 WHERE state = $states";
我试过了
<?
include("connect.php"); // file to connect to db
$states = array(CA,TX);
$states_str = implode(",", $states);
$query="SELECT * FROM table1 WHERE state IN ($states_str)";
$result=mysql_query ($query);
while($row = mysql_fetch_array($result)) {
echo $row['state'];
}
?>
但是我收到此错误消息
Warning: mysql_fetch_array() expects parameter 1 to be resource, boolean given in ... on line 11
这是第11行
while($row = mysql_fetch_array($result)) {
但是，如果我使用此代码，我会得到一个结果，只是为了表明当我切换回旧的常规查询时它可以工作
<?
include("connect.php"); // file to connect to db
$states = array(CA,TX);
$states_str = implode(",", $states);
$query="SELECT * FROM table1 WHERE state = 'CA'";
$result=mysql_query ($query);
while($row = mysql_fetch_array($result)) {
echo $row['state'];
}
?>
但是后来我绕过了内爆（）的东西，我在其他地方找到了它，但它对我不起作用
显然，查询失败是因为字符串值没有用引号引起来。
尝试：
$states = array('CA','TX');
$states_str = implode("','", $states);
$query="SELECT * FROM table1 WHERE state IN ('$states_str')";
<?php
include('../testi.inc');
$states = array('CA','TX');
$states_str = implode("','", $states);
$query = "SELECT * FROM states WHERE state IN ('$states_str')";
$result=mysqli_query ($db,$query) or die(mysql_error());
while($row = mysqli_fetch_assoc($result)) {
echo $row['state'];
}
?>
或类似的东西
<?
include("connect.php"); // file to connect to db
$states = array('CA','TX');
$states_str = implode("','", $states);
$states_str = rtrim($states_str,",'");
$query="SELECT * FROM table1 WHERE state IN ('$states_str')";
$result=mysql_query ($query);
if($result){
while($row = mysql_fetch_array($result)) {
echo $row['state'];
}
}
?>
那是因为您的查询失败了，所以首先检查它