[英]php variable in javascript variable jqgrid integration
我有一個php函數從mysql數據庫中獲取一些數據:
function get_persons() {
$result=dbquery("SELECT id,name,surname FROM ".DB_PERSONS." ORDER BY id ASC");
while ($person = dbarray($result)){
$id=$person['id'];
echo "<option value='$id'>$person['name'] - $person['surname']</option>";
}
}
而且我有一個JavaScript代碼:
$edit_buttonoptions = array("#pager",
array("title"=>"Takım Seç","buttonicon"=>"ui-icon-mail-open","caption"=>"Takım Seç", "onClickButton"=>"js:
var jspersonvar;
jspersonvar = "<?php get_persons();?>"
function(){
var selr = jQuery('#grid').jqGrid('getGridParam','selrow');
var rowData = jQuery('#grid').jqGrid('getRowData', selr);
var kelr = jQuery('#grid').jqGrid('getCell', selr, 'bilinen_adi');
var fotograf = jQuery('#grid').jqGrid('getCell', selr, 'logo');
if(selr)
miktar=miktar+1,
pp='takimlar',
nm=pp + miktar,
jQuery('#takimlar').append('<div id=' + nm + '>'),
jQuery('#'+ nm +'').append('<div style=text-align:center;>'+fotograf+'<input type=hidden name=takim'+ miktar +' value=' + selr + ' />'),
jQuery('#'+ nm +'').append('<select name=sezon'+ miktar +' id=sezon'+ miktar +'>'+jspersonvar+'</select>'),
jQuery('#'+ nm +'').append('' + kelr + ''),
jQuery('#'+ nm +'').append('</div>'),
jQuery('#'+ nm +'').append('</div><br />'),
alert('' + kelr + ' Takımı Seçildi')
else
alert('Lütfen Bir Takım Seçiniz!')
return false;
}"),
);
$grid->callGridMethod("#grid", "navButtonAdd", $edit_buttonoptions);
它不起作用,我已經嘗試了幾件事,但是沒有運氣...
您需要將變量放在引號中,因為服務器呈現頁面時它將如下所示:
jspersonvar = <option value='$id'>......
當您在雙引號周圍jspersonvar
引號時, jspersonvar
將被視為字符串,您可以使用它來做。
jspersonvar = "<?php get_persons();?>"
請注意,數據庫中顯示的任何"
都會破壞您的JavaScript。
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