Question

我正在尝试显示我创建的表并在其中插入新值。另外，两者都有效。但是缺少了一些东西。

在PHP代码下面：

<?php

$dbhost = "host";
$dbuser = "user";
$dbpass = "passowrd";
$dbname = "dbname";

//Connect to MySQL Server
$link = mysql_connect($dbhost, $dbuser, $dbpass);

//Select Database
mysql_select_db($dbname) or die(mysql_error());




// Retrieve data from Query String
$age = $_GET['age'];
$sex = $_GET['sex'];
$wpm = $_GET['wpm'];

// Escape User Input to help prevent SQL Injection
$age = mysql_real_escape_string($age);
$sex = mysql_real_escape_string($sex);
$wpm = mysql_real_escape_string($wpm);

//这两条线不能工作

$query = "INSERT INTO dbname (name, age) VALUES ('mate', 19);";
$query .= "SELECT * FROM dbname";


//Execute query
$qry_result = mysql_query($query) or die(mysql_error());

//Build Result String
$display_string = "<table>";
$display_string .= "<tr>";
$display_string .= "<th>Name</th>";
$display_string .= "<th>Age</th>";
$display_string .= "<th>Sex</th>";
$display_string .= "<th>WPM</th>";
$display_string .= "</tr>";

// Insert a new row in the table for each person returned
while($row = mysql_fetch_array($qry_result)){
   $display_string .= "<tr>";
   $display_string .= "<td>$row[name]</td>";
   $display_string .= "<td>$row[age]</td>";
   $display_string .= "<td>$row[sex]</td>";
   $display_string .= "<td>$row[wpm]</td>";
   $display_string .= "</tr>";
}

echo "Query: " . $query . "<br />";
$display_string .= "</table>";

echo $display_string;

?>

问题是如果我在这里只写一行： $ query =＆＃34; select * from dbname＆＃34 ;; 它工作正常。

两条线有什么问题？

多个查询PHP7 - 找到错误

0 个答案: