[英]Can't get results from mySQL query to display in PHP
I am very new to PHP and can't seem to use mySQL data at all from PHP. 我对PHP非常陌生,似乎根本无法使用PHP中的mySQL数据。 The SQL query I wrote works fine in phpMyAdmin when I run it in the SQL editor, but the most I am able to get from the code is the following from var_dump.
当我在SQL编辑器中运行phpMyAdmin时,我编写的SQL查询工作正常,但是从代码中获得的最多信息是从var_dump中获得的以下信息。 Nothing else displays anything at all.
什么都没有显示。
resource(3) of type (mysql result)
Any help at all would be greatly appreciated!! 任何帮助将不胜感激!
<?php
ini_set(‘display_errors’,1);
error_reporting(E_ALL|E_STRICT);
$con = mysql_connect("localhost","XXXXXXXX","XXXXXXX");
if (!$con) {
die('Could not connect: ' . mysql_error());
}
mysql_select_db("sharetrader", $con);
$sharelist = mysql_query("SELECT DISTINCT tblStocks.stockSymbol, tblShareData.lookupDate FROM tblStocks LEFT JOIN tblShareData ON tblShareData.tickerCode = tblStocks.stockSymbol ORDER BY tblShareData.lookupDate ASC LIMIT 0 , 30");
if (!$sharelist) {
die('Invalid query: ' . mysql_error());
}
var_dump($sharelist);
while ($row = mysql_fetch_array($sharelist)) {
echo $row['tblStocks.stockSymbol'];
}
mysql_close($con);
?>
I am very new to PHP
我是PHP的新手
But you're making great progress - just missing some of the finer points. 但是您正在取得长足进步-只是缺少了一些更好的观点。
try: 尝试:
while ($row = mysql_fetch_array($sharelist)) {
var_dump($row);
}
...and it should be obvious what's happening. ...而且很明显发生了什么事。
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