在php中从数据库获取结果,并将其作为循环输出,以便在javascript中显示
我有这段代码,我正在处理,我正在尝试输出:在php中从数据库获取结果,并将其作为循环输出,以便在javascript中显示,php,javascript,mysql,Php,Javascript,Mysql,我有这段代码,我正在处理,我正在尝试输出: { title:"<?php echo $sender_fullname; ?>", mp3:"link", }, 这是一个循环,我应该怎么做才能输出相同的结果 while( $row = mysql_fetch_assoc($result)) { $sender = $row['sender']; $sender_name_query = mysql_query("SEL
{
title:"<?php echo $sender_fullname; ?>",
mp3:"link",
},
这是一个循环,我应该怎么做才能输出相同的结果
while(
$row = mysql_fetch_assoc($result)) {
$sender = $row['sender'];
$sender_name_query = mysql_query("SELECT fullname FROM users WHERE id = '$sender'");
$sender_name = mysql_fetch_object($sender_name_query);
$sender_fullname = $sender_name->fullname;
echo '{<br/>title:".'$sender_fullname'.",<br/>mp3:"link",<br/>},';
}
while(
$row=mysql\u fetch\u assoc($result)){
$sender=$row['sender'];
$sender\u name\u query=mysql\u query(“从id为“$sender”的用户中选择fullname”);
$sender\u name=mysql\u fetch\u对象($sender\u name\u query);
$sender\u fullname=$sender\u name->fullname;
回音“{
标题:“.$sender_fullname.”,
mp3:“link”,
,”;
}
您可以执行以下操作:
while(
$row = mysql_fetch_assoc($result)) {
$sender = $row['sender'];
$sender_name_query = mysql_query("SELECT fullname FROM users WHERE id = '$sender'");
$sender_name = mysql_fetch_object($sender_name_query);
$sender_fullname = $sender_name->fullname;
echo '{<br/>title:".'$sender_fullname'.",<br/>mp3:"link",<br/>},';
print "<script>alert(\"$sender_fillname\")</script>";
}
while(
$row=mysql\u fetch\u assoc($result)){
$sender=$row['sender'];
$sender\u name\u query=mysql\u query(“从id为“$sender”的用户中选择fullname”);
$sender\u name=mysql\u fetch\u对象($sender\u name\u query);
$sender\u fullname=$sender\u name->fullname;
回音“{
标题:“.$sender_fullname.”,
mp3:“link”,
,”;
打印“警报(\“$sender\u fillname\”);
}
然后,脚本标记中的代码会像javascript代码一样运行,如果您想将php变量的值放入javascript变量中,您可以执行以下操作:
<?
$mivar = "hola mundo";
print "<script>";
print "var mivar = \"$mivar\"";
print "</script>";
?>
将php转换为javascript的最佳方法是通过php的
$rows=array();
而($row=mysql\u fetch\u assoc($result)){
$sender=$row['sender'];
$sender\u name\u query=mysql\u query(“从id为“$sender”的用户中选择fullname”);
$sender\u name=mysql\u fetch\u对象($sender\u name\u query);
$sender\u fullname=$sender\u name->fullname;
$row['sender']=$sender\u全名;
$rows[]=$row;
}
回声“;
echo“var rows=”。json_编码($rows)。";";
回声“;
错误是由最后一行引起的:
echo '{<br/>title:".'$sender_fullname'.",<br/>mp3:"link",<br/>},';
echo'{
标题:“.$sender_fullname.”,
mp3:“link”,
;
应该是这样的:
echo '{<br/>title:"' . $sender_fullname . '",<br/>mp3:"link",<br/>},';
echo'{
标题:“.$sender_fullname.”,
mp3:“link”,
;
注1 你可以这样写:
$sender_fullname = $sender_name->fullname;
echo '{<br/>title:"' . $sender_fullname . '",<br/>mp3:"link",<br/>},';
$sender\u fullname=$sender\u name->fullname;
回显“{
标题:”.$sender_fullname.”,
mp3:“链接”,
,”;
简单地说:
echo '{<br/>title:"' . $sender_name->fullname. '",<br/>mp3:"link",<br/>},';
echo'{
标题:“.$sender_name->fullname.”,
mp3:“link”,
;
注2 。它们不再得到维护。看到了吗?相反,学习,并使用,或-将帮助您决定哪一个。如果您选择PDO
从中,您的问题是什么?现在我得到了这个错误解析错误:语法错误,意外的T_变量,预期为','或';'在里面在第34行,`while($row=mysql\u fetch\u assoc($result)){$sender=$row['sender'];$sender\u name\u query=mysql\u fetch\u object($sender\u name\u query);$sender\u fullname=$sender\u name->fullname;echo“{
title:$sender\u fullname.”,
mp3:“link”,
,
`第34行
echo'{
标题:“.$sender_fullname.”,
mp3:“link”,
代码>我已经添加了一个导致错误的答案。
$sender_fullname = $sender_name->fullname;
echo '{<br/>title:"' . $sender_fullname . '",<br/>mp3:"link",<br/>},';
echo '{<br/>title:"' . $sender_name->fullname. '",<br/>mp3:"link",<br/>},';