通过从Facileforms拉取数据来创建选择列表时出现语法错误

Question

Facileforms对记录id使用#_records，对提交的数据使用#_subrecords。所以，我正在学习内部连接。但是，我得到了一个错误。

错误消息是...“解析错误:语法错误，第16行的/home/snbrown/homestest.dreamhosters.com/components/com_breezingforms/facileforms.process.php(1227)：eval()'d代码中出现意外的T_CONSTANT_ENCAPSED_STRING”

密码是..。

<?php

$user = JFactory::getUser()->get('id');

JFactory::getDBO()->setQuery "SELECT 
#_facileforms_records.id, 
#_facileforms_subrecords.afname, 
#_facileforms_subrecords.alname,
#_facileforms_subrecords.awebsite, 
#_facileforms_subrecords.aphone, 
#_facileforms_subrecords.aemail, 
#_facileforms_subrecords.abrokerage
FROM #_facileforms_records;
INNER JOIN #_facileforms_subrecords;
ON #_facileforms_records.id=#_facileforms_subrecords.record;
WHERE (#_facileforms_subrecords.name = acustomerid AND value = "$userid");
AND (#_facileforms_subrecords.name = "formid AND value = 4)";

?>

当然，作为一个新加入的人，看起来我似乎已经超出了我的承受能力。任何帮助都将不胜感激。

Answer 1

你有一些语法错误，在这里很难解释..

您可以尝试以下fixed代码

<?php

$user = JFactory::getUser()->get('id');

JFactory::getDBO()->setQuery("SELECT 
#_facileforms_records.id, 
#_facileforms_subrecords.afname, 
#_facileforms_subrecords.alname,
#_facileforms_subrecords.awebsite, 
#_facileforms_subrecords.aphone, 
#_facileforms_subrecords.aemail, 
#_facileforms_subrecords.abrokerage
FROM #_facileforms_records;
INNER JOIN #_facileforms_subrecords;
ON #_facileforms_records.id=#_facileforms_subrecords.record;
WHERE (#_facileforms_subrecords.name = acustomerid AND value = '$userid')
AND (#_facileforms_subrecords.name = 'formid' AND value = 4);");
?>

Answer 2

您更正的代码块如下所示。

JFactory::getDBO()->setQuery("SELECT 
#_facileforms_records.id, 
#_facileforms_subrecords.afname, 
#_facileforms_subrecords.alname,
#_facileforms_subrecords.awebsite, 
#_facileforms_subrecords.aphone, 
#_facileforms_subrecords.aemail, 
#_facileforms_subrecords.abrokerage
FROM #_facileforms_records;
INNER JOIN #_facileforms_subrecords;
ON #_facileforms_records.id=#_facileforms_subrecords.record;
WHERE (#_facileforms_subrecords.name = acustomerid AND value = \"$userid\")
AND (#_facileforms_subrecords.name = \"formid\" AND value = 4);");

Answer 3

好吧，有些人已经发布了修复的代码，但他们中没有一个人解释问题是什么，所以我会。这里的问题是您使用双引号将查询字符串括起来，但您也试图在查询中使用双引号。如果您有以下字符串

$myString = "this is a string with "quotes" inside of it";

PHP会认为字符串以with"结尾，而quotes是一个T_CONSTANT_ENCAPSED_STRING (使用define函数定义的常量的标识符)，因此它将抛出一个语法错误。

您可以使用许多方法来消除该问题。在@Roopendra回答中，他用\转义引号，使其成为字符串的一部分，结果是

在@Shankar Damodaran中，他将双引号替换为单引号，因此不会产生混淆。