从mysql中取一个有内容的字段打印出来,却报错了解决办法
Jun 13, 2016 pm 01:23 PM
echo
mysql
quot
从mysql中取一个有内容的字段打印出来,却报错了
打算从MYsql里面取一个字段出来试试nl2br 函数的效果,可是报错了。
写法(目前,数据库中仅一条记录):
- PHP code
<!-- Code highlighting produced by Actipro CodeHighlighter (freeware) http://www.CodeHighlighter.com/ -->$result = mysql_query("SELECT * FROM Product limit 40"); while($row = mysql_fetch_array($result)) { echo nl2br($row['re_mark']); //就是这行报错了 }
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re_mark字段类型是:text
如果把re_mark改成id字段,倒是不报错。
- PHP code
<!-- Code highlighting produced by Actipro CodeHighlighter (freeware) http://www.CodeHighlighter.com/ -->$result = mysql_query("SELECT * FROM Product limit 40"); while($row = mysql_fetch_array($result)) { echo "<img src="%24row%5B'pic'%5D" alt=" 从mysql中取一个有内容的字段打印出来,却报错了解决办法 " >";//这样写也报错 }
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报错内容:
Parse error: syntax error, unexpected T_ENCAPSED_AND_WHITESPACE, expecting T_STRING or T_VARIABLE or T_NUM_STRING in D:\xampp\htdocs\qry.php on line 9
------解决方案--------------------
var_dump($row['pic']); //贴出结果
------解决方案--------------------
------解决方案--------------------
echo "
这是必然的,因为你违反了php的书写约定
这样都是可以的:
echo "
echo "
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