标签: syntax-error

我正在使用立体视觉的源代码,它给出了一个错误

1>  StereoMain.cpp
1>c:\opencv2.2\include\opencv2\highgui\highgui_c.h(171): error C2059: syntax error : 'constant'
1>c:\opencv2.2\include\opencv2\highgui\highgui_c.h(171): error C3805: 'constant': unexpected token, expected either '}' or a ','
1>  StereoGrabber.cpp
1>c:\opencv2.2\include\opencv2\flann\logger.h(66): warning C4996: 'fopen': This function or variable may be unsafe. Consider using fopen_s instead. To disable deprecation, use _CRT_SECURE_NO_WARNINGS. See online help for details.
1>          e:\program files\microsoft visual studio 10.0\vc\include\stdio.h(234) : see declaration of 'fopen'
1>c:\opencv2.2\include\opencv2\highgui\highgui_c.h(171): error C2059: syntax error : 'constant'
1>c:\opencv2.2\include\opencv2\highgui\highgui_c.h(171): error C3805: 'constant': unexpected token, expected either '}' or a ','
1> …

我使用以下代码获得了一个Unxepected Identifier编辑器.我整天都在排它故障,什么都没有.希望一双新鲜的眼睛可以发现我的错误.我正在尝试使用jQuery UI来设置日期选择器,然后获取并操纵日期.更改选择器上的日期应通过ajax更改与该日期关联的页面上的图像.

$(document).ready(function(){
// Datepicker
$('#datepicker').datepicker({
    dateFormat: 'yy-mm-dd',
    inline: true,
    minDate: new Date(2012, 06 - 1, 1),
    maxDate:new Date(2012, 09 - 1, 31),
    onSelect: function(){
       var day1 = ($("#datepicker").datepicker('getDate').getDate()).toString().replace(/(^.$)/,"0$1");              
       var month1 = ($("#datepicker").datepicker('getDate').getMonth() + 1).toString().replace(/(^.$)/,"0$1");             
        var year1 = $("#datepicker").datepicker('getDate').getFullYear();
        var fullDate = year1 + "/" + month1 + "/" + day1;
        var dashDate = year1 + "-" + month1 + "-" + day1;
        var str_output = "<a id=\"single_image\" href=\"http://www.lasalle.edu/150/dayinhistory/" + fullDate + ".jpg\" title=\"\"><img src=\"http://www.lasalle.edu/scripts/timthumb/timthumb.php?src=http://www.lasalle.edu/150/dayinhistory/" + fullDate + ".jpg&w=560&h=350&zc=1&a=t\">"; …

我有"通常"预期的缩进块.所有的缩进都是正确的.我已经在各种编辑器中打开了我的脚本,并且没有任何问题可能会导致错误,或隐藏由标签引起的空白.

如果有人能对这个问题有所了解,真的很感激.

这是导致问题的脚本的一部分:

def findCSVs():
'''
looks into a 'tocom_data' subdirectory, finds 'tocomxxx.csv' files,
retuns a sorted list of filenames that conform: begins with TOCOM, ends in .csv
'''
    csvlist = []
    datadir=os.path.join('.','tocom_data')
    flist = os.listdir(datadir)
    for fname in flist:
        fsplit = fname.split('.')
        if len(fsplit)>1:
            if fsplit[1]=="csv" and fname[0:5]=="TOCOM":
                completeFname= os.path.join(datadir,fname)
                csvlist.append(completeFname)
                csvlist.sort()
    return csvlist

Python期望该行有一个缩进块 if len(fsplit)>1:

非常感激

何塞

这是代码在Perl上给我错误.

...
else if (exists($framename{$presFrame}) && (($framename{$presFrame}) < = $j))
...

这framename是一个哈希,presFrame是一个关键存在于framename

编辑:

在正确答案中提到的Perl中实现if/else语法有点不同.

我在firebug中收到此错误:

SyntaxError: missing ) after argument list

从这段代码:

table +=
'<tbody>'+
  '<tr>'+
  '<td>'+this.zona+'</td>' +
  '<td>'+this.rruga+'<br/><div id="dialog-'+this.id+'" style= "margin-left:1px; width:340px; display:none;height:180px"></div></td>' +
  '<td>'+this.dita+'</td>' +
  '<td><img src="images/map.jpg"  id = "mapicon" onClick="show(this.rruga, this.id);"></td>' +
'<tr>'+
'<tbody>';

我不太确定)我的代码中缺少哪个括号.

任何人都可以帮我调试这个问题吗？这是我想从cron运行的php脚本的一部分.我一直在看它,但无法弄清楚出了什么问题.以下行抛出PHP错误"PHP Parse错误:语法错误,意外的T_LNUMBER在......行..."

$sqlreadstmt = $db->query("
select distinct 
rsdata.rs_variant,
rsdata.weeknumber,
rsdata.rs_moduleid,
rsdata.rs_objectidentifier,
trdata.tr_objectidentifier,
trdata.tr_testversion,
trdata.tr_plannedgate,
trdata.tr_vnvmethod,
testatussequence.tesequencenr,
trdata.tr_testexecutionstatus 
from rsdata,trdata,module,variantgates,testatussequence 
where rsdata.weeknumber="1339" 
and rsdata.rs_moduleid = module.moduleid 
and rsdata.weeknumber = trdata.weeknumber 
and rsdata.rs_reviewstatus != "Obsolete" 
and rsdata.rs_reviewstatus != "Rejected" 
and rsdata.rs_introbjectidentifieralllevels != "" 
and rsdata.rs_introbjectidentifieralllevels != "Unknown" 
and find_in_set(trdata.tr_objectidentifier, (select rsdata.rs_introbjectidentifieralllevels)) 
and trdata.tr_plannedgate != "" 
and trdata.tr_plannedgate != "Unknown" 
and trdata.tr_plannedgate = variantgates.gate 
and trdata.tr_variant = variantgates.variant 
and trdata.tr_testexecutionstatus = testatussequence.testatus 
order by 
rs_variant ASC,
weeknumber ASC,
rs_moduleid ASC,
rs_objectidentifier ASC,
tr_testversion …

我对编码很新,我不断收到这个错误,我真的需要帮助.这是我的代码:

public String getArmorTexture(ItemStack stack, Entity entity, int slot, String type){
    if (stack.getItem() == halo.TitaniumHelmet || stack.getItem() == halo.TitaniumChestplate || stack.getItem() == halo.TitaniumBoots) {
        return "halo:textures/models/armor/Titanium1.png";
    }
    if (stack.getItem() == halo.TitaniumLeggings); {
        return "halo:textures/models/armor/Titanium_layar_2.png";
    } else { //<------ Syntax error on token "else", delete this token
        return null; 
    }

我有:

if($^O eq 'MSWin32'){
  export WINDOWS=1
else{
  export UNIX=1
}

=begin WINDOWS
  use feature qw(switch);
=cut
=begin UNIX
  use Switch;
=cut

我得到:

C:\ build.pl第6行的语法错误,"else"附近BEGIN错误后不安全 - 编译在C:\ build.pl第17行中止.

我在下面的代码中遇到错误.解析错误:语法错误,第7行意外"大小"(T_STRING).请帮助

<html>
<head>
</head>
<body>
<?php
function fontWrap( $txt, $size ) {
print “<font size=\”$size\”
face=\”Helvetica,Arial,Sans-Serif\”>
$txt</font>”;
}
fontWrap(“A heading<br>”,5);
fontWrap(“some body text<br>”,3);
fontWrap(“some more body text<BR>”,3);
fontWrap(“yet more body text<BR>”,3);
?>
</body>
</html>

这段代码出了什么问题？尝试导入此sql时收到该消息:

#1064 - 您的SQL语法有错误; 查看与您的MySQL服务器版本对应的手册,以便在''codigo'int附近使用正确的语法(11)无符号NOT NULL auto_increment,'razao_social'varchar(11)N'在第2行

CREATE TABLE IF NOT EXISTS `char` (
'codigo' int(11) unsigned NOT NULL auto_increment,
'razao_social' varchar(11) NOT NULL DEFAULT '',
'fantasia' varchar(11) NOT NULL DEFAULT '',
'data_de_cadastro' datetime NOT NULL DEFAULT '',
'cep' varchar(11) NOT NULL DEFAULT '',
'logradouro' varchar(11) NOT NULL DEFAULT '',
'numero' varchar(11) NOT NULL DEFAULT '',
'complemento' varchar(11) NOT NULL DEFAULT '',
'bairro' varchar(11) NOT NULL DEFAULT '',
'cidade' varchar(11) NOT NULL DEFAULT '',
'uf' varchar(11) NOT NULL DEFAULT '', …