Question

SQL查询很好，但以下代码不返回任何值。我需要显示查询中的单个值。

$clientId=$row["clientid"];
        $sql2="select cname from Clients where clientid='$clientId'";
        //$result = $conn->query($sql2);
        $result = mysql_query($sql2);
        echo $result;

Answer 1

为了使用你需要从中获取的结果。 E.g：

$result = mysql_query($sql2);
$row = mysql_fetch_array($result);
$cname = $row[0];

Answer 2

$clientId=$row["clientid"];
        $sql2="select cname from Clients where clientid='$clientId'";
        //$result = $conn->query($sql2);
        $result = mysql_query($sql2);
        foreach($result->result() as $row){
         echo $row->cname;
}

Answer 3

尝试这种方式并使用（`）标记来修复您的SQL查询错误，但首先请参阅此弃用API的文档，而不是使用 PDO 或 MYSQLi .see mysql_query

$clientId=$row["clientid"];
$sql2="select `cname` from Clients where clientid='$clientId'";
$result = mysql_query($sql2);
$row = mysql_fetch_assoc($result);
//use column name,as you fetch result as associative array
$canme = $row['cname']; 
echo $canme;

N.B：使用PDO或MYSQLi，用于修复您的SQL INJECTION相关问题的预备语句

Answer 4

这是你应该如何使用mysqli

来完成的

$query = 'SELECT cname FROM Clients WHERE clientid = ?';

if($stmt = $mysqli->prepare($query)){
    $stmt->bind_param('i', $_POST['clientid']);
    $stmt->execute();
    $stmt->bind_result($cname);
    $stmt->fetch();

    echo $cname; // (optional, prints it out)
}

Answer 5

不推荐使用mysql

你正在应用错误的方法。

您的代码应如下所示：

<?php
$con=mysqli_connect("localhost","my_user","my_password","my_db");
// Check connection
if (mysqli_connect_errno())
{
  echo "Failed to connect to MySQL: " . mysqli_connect_error();
  }

$sql="SELECT Lastname,Age FROM Persons ORDER BY Lastname";
$result=mysqli_query($con,$sql)

// Numeric array
$row=mysqli_fetch_array($result,MYSQLI_NUM);
printf ("%s (%s)\n",$row[0],$row[1]);

// Associative array
$row=mysqli_fetch_array($result,MYSQLI_ASSOC);
printf ("%s (%s)\n",$row["Lastname"],$row["Age"]);

// Free result set
mysqli_free_result($result);

mysqli_close($con);
    ?>

还准备好更多＆gt;＆gt; http://www.w3schools.com/php/func_mysqli_fetch_array.asp

如何从PHP中的MySQL查询中获取单个值

5 个答案: