我有以下部分代码可以在本地计算机上运行,但不能在服务器上运行。
当我在服务器上执行它时会收到错误" Uncaught SyntaxError:Unexpected token;"
<?php
$checkValue = array();
$i= 0;
while ($row = mysql_fetch_array($searchdetails)) {
$checkValue[$i][] = $row['ID'];
$checkValue[$i][] = $row['1'];
$checkValue[$i][] = $row['2'];
$checkValue[$i][] = $row['3'];
if(@$row['4'] == "NULL" || @$row['4'] == NULL) {
$checkValue[$i][] = "NULL";
}else{
$sql = "SELECT * FROM Class WHERE ID_Class =".@$row['4'];
$data = mysql_query( $sql );
$class = mysql_fetch_assoc($data);
$checkValue[$i][] = $class['Class'];
}
if(@$row['5'] == "NULL" || @$row['5'] == NULL) {
$checkValue[$i][] = "NULL";
}else{
$sql = "SELECT * FROM Place WHERE ID_Place =".@$row['5'];
$data = mysql_query( $sql );
while($row1 = mysql_fetch_array($data)) {
$checkValue[$i][] = $row1['1'];
}
}
$i++;
}
?>
<script>
var grid;
var columns = [
{ id: "switch", name: "switch", field: "switch", formatter: imageFormatter, sortable: true },
{ id: "college", name: "college", field: "college", width: 50 },
{ id: "Number", name: "Number", field: "Number", width: 50 },
{ id: "Class", name: "Class", field: "Class",sortable: true },
{ id: "Place", name: "Place", field: "Place", width: 40 },
];
$(function () {
var MS_PER_DAY = 24 * 60 * 60 * 1000;
var data = [];
**var listdata = <?php echo json_encode($checkValue);?>;**
console.log(listdata);
for (var i = 0; i < listdata.length; i++) {
data[i] = {
switch: listdata[i][0],
college: listdata[i]['1'],
Number: listdata[i]['2'],
Strength: listdata[i]['3'],
Place: listdata[i]['4'],
};
}
})
</script>
我在Stackoverflow中搜索不同的答案后尝试了这个。
var listdata = <?php echo json_encode($checkValue)?>;
但它仍然没有用。
在这里阻止它的问题可能是什么。对此有任何帮助,我们非常感激!!
编辑:我确定了原因,表类中有一些西班牙语字符,因此它不显示数据。如何克服这些角色?