PHP报价问题

我有一个名为user的文本字段和一个submit按钮，该按钮显示一个Json字符串，其中包含数据库中具有该id的用户的详细信息。我有以下代码：

$UserID = mysql_real_escape_string($_POST['User']);

$UserCoords_Query  = "Select Accuracy, Longitude, Latitude, Timestamp 
                      FROM UserDetails
                      WHERE UserId =".$UserID;

$UserCoords_result = mysql_query($UserCoords_Query);

if (mysql_num_rows($UserCoords_result) == 0) {
    echo "There are no users with an id of ". $UserID;
}

但我得到了一个错误：

$UserID = mysql_real_escape_string($_POST['User']);

$UserCoords_Query  = "Select Accuracy, Longitude, Latitude, Timestamp 
                      FROM UserDetails
                      WHERE UserId =".$UserID;

$UserCoords_result = mysql_query($UserCoords_Query);

if (mysql_num_rows($UserCoords_result) == 0) {
    echo "There are no users with an id of ". $UserID;
}

警告：mysql\u num\u rows（）希望参数1是第29行/home/roseanne/public\u html/display\u Coords.php中给出的资源布尔值。

$UserID = mysql_real_escape_string($_POST['User']);

$UserCoords_Query  = "Select Accuracy, Longitude, Latitude, Timestamp 
                      FROM UserDetails
                      WHERE UserId =".$UserID;

$UserCoords_result = mysql_query($UserCoords_Query);

if (mysql_num_rows($UserCoords_result) == 0) {
    echo "There are no users with an id of ". $UserID;
}

---

我最近将数据库中的userID类型从int改为text。它作为一个int起作用，但自更改后就没有了。有人能看到引号问题吗？

Timestamp

是mysql中的保留字，请在其周围使用

`

$UserID = mysql_real_escape_string($_POST['User']);

$UserCoords_Query  = "Select Accuracy, Longitude, Latitude, Timestamp 
                      FROM UserDetails
                      WHERE UserId =".$UserID;

$UserCoords_result = mysql_query($UserCoords_Query);

if (mysql_num_rows($UserCoords_result) == 0) {
    echo "There are no users with an id of ". $UserID;
}

$UserCoords_Query  = "Select `Accuracy`, `Longitude`, `Latitude`, `Timestamp`
                  FROM UserDetails
                  WHERE UserId = '".$UserID."'";

---

mysql\u query

失败时返回false，因此您必须检查它。在您的例子中，它返回了

false

——您的查询失败，但您仍然将该

false

值传递到函数

mysql\u num\u rows

，这引发了一个错误

$UserID = mysql_real_escape_string($_POST['User']);

$UserCoords_Query  = "Select Accuracy, Longitude, Latitude, Timestamp 
                      FROM UserDetails
                      WHERE UserId =".$UserID;

$UserCoords_result = mysql_query($UserCoords_Query);

if (mysql_num_rows($UserCoords_result) == 0) {
    echo "There are no users with an id of ". $UserID;
}

另外，您应该使用

mysqli.*

，因为

mysql.*

已被弃用。

$UserID = mysql_real_escape_string($_POST['User']);

$UserCoords_Query  = "Select Accuracy, Longitude, Latitude, Timestamp 
                      FROM UserDetails
                      WHERE UserId =".$UserID;

$UserCoords_result = mysql_query($UserCoords_Query);

if (mysql_num_rows($UserCoords_result) == 0) {
    echo "There are no users with an id of ". $UserID;
}

$UserID = mysql_real_escape_string($_POST['User']);

$UserCoords_Query  = "Select `Accuracy`, `Longitude`, `Latitude`, `Timestamp` 
                      FROM UserDetails
                      WHERE UserId =".$UserID;

$UserCoords_result = mysql_query($UserCoords_Query);

if ($UserCoords_result && mysql_num_rows($UserCoords_result) >= 1) {
  //Fetch results
} else {
  echo "There are no users with an id of ". $UserID;
}

---

$UserCoords\u result

未被设置为函数

mysql\u num\u rows（）

的有效资源。这可能是因为mysql_查询在您没有首先创建到数据库的连接的情况下试图运行。我需要看更多的代码才能确定

$UserID = mysql_real_escape_string($_POST['User']);

$UserCoords_Query  = "Select Accuracy, Longitude, Latitude, Timestamp 
                      FROM UserDetails
                      WHERE UserId =".$UserID;

$UserCoords_result = mysql_query($UserCoords_Query);

if (mysql_num_rows($UserCoords_result) == 0) {
    echo "There are no users with an id of ". $UserID;
}

另外，在您开发此功能的同时，我建议您迁移到mysqli或DBO，因为从PHP5.5开始，mysql_uu函数正在贬值

$UserID = mysql_real_escape_string($_POST['User']);

$UserCoords_Query  = "Select Accuracy, Longitude, Latitude, Timestamp 
                      FROM UserDetails
                      WHERE UserId =".$UserID;

$UserCoords_result = mysql_query($UserCoords_Query);

if (mysql_num_rows($UserCoords_result) == 0) {
    echo "There are no users with an id of ". $UserID;
}

---

：）

您确定已正确连接并选择了数据库吗？您正在使用并且应该使用。@Yousf是的，仍然连接！我认为这不是问题所在，因为当我在phpmyadmin中运行查询时，它是有效的。这是一个引用问题：/@user2363025这就是问题所在，因为您的查询失败。@user2363025请参阅带引号的编辑查询。我已编写了连接到数据库的代码，在更改数据类型之前，查询已运行。我认为查询中的$UserId需要加引号，但我不知道如何使用这种语法。谢谢你的建议！：）排序完毕后，我需要更改为：$UserCoords_Query=“从UserDetails中选择精度、经度、纬度、时间戳，其中UserId=”””。$UserId。“”；

$UserID = mysql_real_escape_string($_POST['User']);

$UserCoords_Query  = "Select Accuracy, Longitude, Latitude, Timestamp 
                      FROM UserDetails
                      WHERE UserId =".$UserID;

$UserCoords_result = mysql_query($UserCoords_Query);

if (mysql_num_rows($UserCoords_result) == 0) {
    echo "There are no users with an id of ". $UserID;
}