Question

我正在尝试将我的下面的数据库字段（名称）的值分配给php变量（名称）。但是我得到的错误。可以帮助

Parse error: syntax error, unexpected T_ENCAPSED_AND_WHITESPACE, expecting T_STRING or   T_VARIABLE or T_NUM_STRING in /home/a4363282/public_html/pune/upload/upload_file.php on line 6

include './connection.php';
$query = "select * from shops WHERE city='pune' AND Ref=1;
$row = mysql_query($query);
$name=$row['sname'];
echo $name;

Answer 1

您在查询结尾时缺少双重报价，更新此，

include './connection.php';
$query = "select * from shops WHERE city='pune' AND Ref=1";
$row = mysql_query($query);
$name=$row['sname'];
echo $name;

Answer 2

请将您的代码更新为MySQLi或PDO。

<?php
include './connection.php';
$query = "select * from shops WHERE city='pune' AND Ref=1";
$result = mysql_query($query);
while($row = mysql_fetch_assoc($result))
{
  echo $row['sname'];
}
?>

Answer 3

避免使用mysql_ *语句。使用 -

更改您的代码

$mysqli = new mysqli("localhost", "my_user", "my_password", "world");


if (mysqli_connect_errno()) {
    printf("Connect failed: %s\n", mysqli_connect_error());
    exit();
}

$query = "SELECT * FROM shops WHERE city=? AND Ref=?";

if ($stmt = $mysqli->prepare($query)) {

    $stmt->bind_param("si", 'pune', 1);

    $stmt->execute();

    $stmt->bind_result($col1, $col2);

    $array = array();
    while ($fetch = $stmt->fetch()) {
        $array[] = $fetch;
    }
    print_r($array);

    $stmt->close();
}

$mysqli->close();

Answer 4

我不知道你要用这个代码做什么，因为它错了

尝试打印$ result将不允许访问资源中的信息

$row = mysql_query($query);

这里$ row是资源所以，必须使用其中一个mysql结果函数。所以你的代码将是

include './connection.php';
$query = "select * from shops WHERE city='pune' AND Ref='1'";
$row = mysql_query($query);

 while($ansArray = mysql_fetch_assoc($row)){

   echo $name = $ansArray['sname'];
 }

将mysql字段值分配给PHP变量

4 个答案: