2017-06-17 72 views
-1

嗨,你可以请帮我找出任何错误在这个特殊的代码,我是新来PHP时我不知道多少,但错误是

解析错误:语法错误,意外 '$ sql_query'(T_VARIABLE)

<?php 

require"init.php"; 

$user_name = "Nikhil123"; 
$user_pass = "Sawant" 

$sql_query = "select name from user_info where username like '$user_name' and userpass like '$user_pass' ;" ; 

$result = mysqli_query($con,$sql_query); 

if(mysqli_num_rows($result)>0) 
{ 
    $row = mysqli_fetch_assoc($result); 
    $name = $row("name"); 
    echo "<h3> HELLO welcome".$name "</h3>"; 
} 
else 
{ 
    echo "No INFO"; 
} 

?> 

回答

2

$user_pass = "Sawant"后添加;。添加."</h3>";echo "<h3> HELLO welcome".$name "</h3>";之间$name

<?php 

require"init.php"; 

$user_name = "Nikhil123"; 
$user_pass = "Sawant"; 

$sql_query = "select name from user_info where username like '$user_name' and userpass like '$user_pass' ;" ; 

$result = mysqli_query($con,$sql_query); 

if(mysqli_num_rows($result)>0) 
{ 
    $row = mysqli_fetch_assoc($result); 
    $name = $row("name"); 
    echo "<h3> HELLO welcome".$name ."</h3>"; 
} 
else 
{ 
    echo "No INFO"; 
} 

?> 
+0

谢谢! ,这只是一个愚蠢的错误 –

-1

使用单引号,而不是 “...” 的SQL - 即 '选择...' ...... “; '是$ var。”'


$ sql_query = '从USER_INFO其中username喜欢选择名称 “和为userpass像 ''。$ USER_NAME。 ”“' $ user_pass。 '”;' ;

+0

这不是问题的根源。 https://stackoverflow.com/questions/11321491/when-to-use-single-quotes-double-quotes-and-backticks-in-mysql – Rickard

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